Coordinate Geometry

We know that,

Abscissa is the x-coordinate and ordinate is the y-coordinate of a point.

(i) The coordinate of the point whose abscissa is 3 and ordinate is -4 is (3, -4).

(ii) The coordinate of the point whose abscissa is $-\frac{3}{2}$ and ordinate is 5 is $\Big(-\frac{3}{2}, 5)$.

(iii) The coordinate of the point whose whose abscissa is $-1\frac{2}{3}$ and ordinate $-2\frac{1}{4}$ is $\Big(-1\frac{2}{3}, -2\frac{1}{4}\Big)$.

(iv) The coordinate of the point whose ordinate is 5 and abscissa is -2 is (-2, 5).

(v) Since, point lies on x-axis. So, ordinate of point will be = 0.

The coordinate of the point whose abscissa is -2 and lies on x-axis is (-2,0).

(vi) Since, point lies on y-axis. So, abscissa of point will be = 0.

The coordinate of the point whose ordinate is $\frac{3}{2}$ and lies on y-axis is $\Big(0, \frac{3}{2}\Big)$.

From figure,

(-3, 5) lies in second quadrant.

(4, -1) lies in fourth quadrant.

(2, 0) lies on x-axis.

(2, 2) lies in first quadrant.

(-3, -6) lies in third quadrant.

Given points are A (0, 2), B (5, 6), C (23, 0), D (0, 23), E (0, -4), F (-6, 0), G ($\sqrt{3}$,0)

(i) If y-coordinate of a point is zero, then the point lies on x-axis.

Hence, C(23, 0), F(-6, 0) and G($\sqrt{3}$, 0) lies on x-axis.

(ii) If x-coordinate of a point is zero, then the point lies on y-axis.

Hence, A(0, 2), D(0, 23) and E(0, -4) lies on y-axis.

The points are shown in the below graph:

From figure,

The co-ordinates of the points are:

Point	Co-ordinates
A	(2, 2)
B	(-3, 0)
C	(-2, -4)
D	(3, -1)
E	(-4, 4)
F	(0, -2)
G	(2, -3)
H	(0, 3)

In point A(2, 2), both x and y coordinates are positive. So it lies in the first quadrant.

In point B(-3, 0), y-coordinate is zero. So it lies on x-axis.

In point C(-2, -4), both x and y coordinates are negative. So it lies in the third quadrant.

In point D(3, -1), x coordinate is positive and y coordinate is negative. So it lies in the fourth quadrant.

The points are shown on the graph below:

The points are shown on the graph below:

(i)

Points (1, 3), (-1, -1) and (-2, -3) lie on a straight line, so they are collinear.

(ii)

Points (1, 2), (2, -1) and (-1, 4) do not lie on a line. So they are non-collinear.

(iii)

Points (0, 1), (2, -2) and $(\frac{2}{3} ,0)$ lie on a straight line, so they are collinear.

The points are shown on the graph below:

Steps of construction :

1. Plot point P(-3, 4) on graph.

2. Draw PM and PN which are perpendiculars to x-axis and y-axis respectively.

From figure,

Coordinates of point M = (-3, 0) and N = (0, 4).

The points are shown on the graph below:

Steps of construction :

1. Plot the points A (1, 2), B (-4, 2), C (-4, -1) and D (1, -1) on graph.

2. Join AB, BC, CD and AD.

From graph,

ABCD is a rectangle.

1 block = 1 unit

AB = 5 units, AD = 3 units

Area of rectangle ABCD = length × breadth

= AB × AD

= 5 × 3

= 15 sq. units.

Hence, ABCD is a rectangle and its area = 15 sq. units.

The points are shown on the graph below:

Steps of construction :

1. Plot the points A(0, 2), B(3, 0), C(0, -2) and D(-3,0) on graph.

2. Join AB, BC, CD and DA. ABCD is a rhombus.

3. Join BD and AC.

BD and AC are the diagonals of the rhombus.

Area of a rhombus = $\frac{1}{2}$ × d₁ × d₂

From graph,

1 block = 1 unit.

AC = 4 units

BD = 6 units.

Area of rhombus ABCD = $\frac{1}{2}$ × BD × AC

= $\frac{1}{2}$ × 6 × 4

= 12 sq. units.

Hence, figure obtained form graph is a rhombus with area = 12 sq. units.

The points are shown on the graph below:

Steps of construction :

1. Plot points A(2, 3), B(-3, 3) and C(-3, -2) on graph.

2. Measure AB.

3. Mark point D such that it is at a distance AB from points A and C.

4. Join AB, BC, CD and DA.

On measuring,

AB = 5 units [As, 1 block = 1 unit]

Area of the square = side × side

Area of the square ABCD = AB × AB

= 5 × 5 = 25 sq. units.

Hence, the coordinates of D = (2, -2) and area of the square is 25 sq. units.

The points are shown on the graph below:

Given,

The rectangle which is 6 units long and 4 units wide is shown in the graph.

Rectangle is in the first quadrant.

Longer side lies on x-axis and one vertex is at origin A(0, 0).

Steps of construction :

1. Mark point A(0, 0).

2. At a distance of 6 units on x-axis mark point B(6, 0).

3. From B draw a line segment parallel to y-axis and mark point C on it at distance of 4 units.

4. From C draw a line segment parallel to x-axis and mark point D on it at distance of 6 units.

5. Join DA.

Hence, coordinates of the rectangle are A(0, 0), B(6, 0), C(6, 4) and D(0, 4).

Given,

Length = 6 units.

Breadth = 3 units

AB = 6 units.

As, 1 block = 1 unit

From graph,

A = (-3, 0)

B = (3, 0)

C = (3, 3)

D = (-3, 3).

Given equilateral triangle OAB.

OA = OB = AB = 2a units.

Draw AD ⊥ OB.

In an equilateral triangle, a perpendicular drawn from one of the vertices to the opposite side bisects the side.

∴ OD = $\frac{1}{2}$ x OB = $\frac{1}{2}$ x 2a = a.

In right angle triangle OAD,

⇒ OA² = OD² + AD²

⇒ (2a)² = a² + AD²

⇒ 4a² = a² + AD²

⇒ AD² = 4a² - a²

⇒ AD² = 3a²

⇒ AD = $\sqrt{3a}$ units.

⇒ AD = $\sqrt{3}$a units.

From graph,

Co-ordinates of O = (0, 0)

Co-ordinates of B = (2a, 0)

As, OD = a units and AD = $\sqrt{3}$a units.

Co-ordinates of A = (a, $\sqrt{3}$a).

Hence, co-ordinates of O = (0, 0), B = (2a, 0) and A = (a, $\sqrt{3}$a).

Given, PQR is an equilateral triangle in which Q(0, 2) and R(0, -2) and O = (0, 0).

Let (x, 0) be the coordinates of P. [As, P lies on x axis, so y-coordinate is zero.]

By distance formula,

$$\begin{aligned}\Rightarrow QR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(0 - 0)^2 + (-2 - 2)^2} \\[1em] = \sqrt{0 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] \Rightarrow OQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(0 - 0)^2 + (2 - 0)^2} \\[1em] = \sqrt{0 + (2)^2} \\[1em] = \sqrt{4} \\[1em] = 2 \text{ units}.\end{aligned}$$

PQ = PR = QR = 4.

OQ = 2

In right angle triangle POQ,

⇒ PQ² = OP² + OQ² [By pythagoras theorem]

⇒ 4² = OP² + 2²

⇒ 16 = OP² + 4

⇒ OP² = 16 - 4

⇒ OP² = 12

⇒ OP = $\sqrt{12} = 2\sqrt{3}$.

Since, P lies on x-axis and OP = $2\sqrt{3}$.

Hence, the coordinates of P are ($2\sqrt{3}$, 0).

The given equation 2x + y + 3 = 0, can be written as :

⇒ y = -2x - 3

When x = 0, y = -2(0) - 3 = -3,

x = 1, y = -2(1) - 3 = -2 - 3 = -5,

x = 2, y = -2(2) - 3 = -4 - 3 = -7.

Table of values :

x	y
0	-3
1	-5
2	-7

Steps of construction :

1. Plot the points (0, -3), (1, -5) and (2, -7) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

The given equation x - 5y - 4 = 0, can be written as :

⇒ 5y = x - 4

⇒ y = $\frac{1}{5}$(x - 4)

When x = -6, y = $\frac{1}{5}(-6 - 4) = -\frac{10}{5}$ = -2,

x = -1, y = $\frac{1}{5}(-1 - 4) = -\frac{5}{5}$ = -1,

x = 4, y = $\frac{1}{5}(4 - 4) = \frac{0}{5}$ = 0.

Table of values :

x	y
-6	-2
-1	-1
4	0

Steps of construction :

1. Plot the points (-6, -2), (-1, -1) and (4, 0) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

The above equation, 3y = 12 - 2x can be written as :

⇒ 3y = 12 - 2x

⇒ y = $\frac{12}{3} - \frac{2x}{3}$

⇒ y = 4 - $\frac{2x}{3}$.

When x = -3, y = 4 - $\frac{(2 \times -3)}{3} = 4 - \frac{-6}{3}$ = 4 - (-2) = 6,

x = 0, y = 4 - $\frac{2 \times 0}{3} = 4 - 0$ = 4,

x = 3, y = 4 - $\frac{2 \times 3}{3}$ = 4 - 2 = 2.

Table of values :

x	y
-3	6
0	4
3	2

Steps of construction :

1. Plot the points (-3, 6), (0, 4) and (3, 2) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

The above equation, 5x + 6y - 30 = 0 can be written as :

⇒ 6y = -5x + 30

⇒ y = $\frac{1}{6}(-5x + 30)$

⇒ y = $-\frac{5x}{6} + 5$.

When x = 0, y = -$\frac{(5 \times 0)}{6} + 5$ = 0 + 5 = 5,

x = 6, y = -$\frac{5 \times 6}{6} + 5$ = -5 + 5 = 0,

x = 12, y = -$\frac{5 \times 12}{6}$ + 5 = -10 + 5 = -5.

Table of values :

x	y
0	5
6	0
12	-5

Steps of construction :

1. Plot the points (0, 5), (6, 0) and (12, -5) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

By formula,

Area of triangle = $\frac{1}{2} \times \text{ base} \times \text{ height}$

From graph,

Base = 6 units, Height = 5 units.

Area = $\frac{1}{2} \times 6 \times 5 = 15$ sq. units.

Hence, the graph of the given equation is shown in the adjoining figure and area of triangle = 15 sq. units.

The above equation, 4x - 3y + 12 = 0 can be written as :

⇒ 3y = 4x + 12

⇒ y = $\frac{4}{3}x + 4$

When x = -6, y = $\frac{4}{3} \times -6 + 4$ = -8 + 4 = -4,

x = -3, y = $\frac{4}{3} \times -3 + 4 =$ -4 + 4 = 0,

x = 0, y = $\frac{4}{3} \times 0 + 4 =$ 0 + 4 = 4.

Table of values :

x	y
-6	-4
-3	0
0	4

Steps of construction :

1. Plot the points (-6, -4), (-3, 0) and (0, 4) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

By formula,

Area of triangle = $\frac{1}{2} \times \text{ base} \times \text{ height}$

From graph,

Base = 3 units, Height = 4 units.

Area = $\frac{1}{2} \times 3 \times 4 = 6$ sq. units.

Hence, the graph of the given equation is shown in the adjoining figure and area of triangle = 6 sq. units.

Equation :

⇒ y = 3x - 4

When x = 0, y = 3(0) - 4 = 0 - 4 = -4,

x = 2, y = 3(2) - 4 = 6 - 4 = 2,

Table of values :

x	y
0	-4
2	2

Steps of construction :

1. Plot the points (0, -4) and (2, 2) on the graph.

2. Connect the two points by a straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

(i) Steps of construction :

1. From point, P (x = -1) draw a line parallel to y-axis touching the graph. Mark the point as Q.

2. From Q draw a line parallel to x-axis, touching y-axis at point R (y = -7).

Hence, y = -7, when x = -1.

(ii) Steps of construction :

1. From point, S (y = 5) draw a line parallel to x-axis touching the graph. Mark the point as T.

2. From T draw a line parallel to y-axis, touching x-axis at point U (x = 3).

Hence, x = 3, when y = 5.

Steps of construction :

1. Plot the points (4, 0) and (0, 3) on graph.

2. Connect the points through a straight line.

3. Take a point Q (y = 1.5) and draw a line parallel to x-axis touching the graph at point R.

4. From point R draw a line parallel to y-axis and touching x-axis at point S (x = 2).

From graph R = (2, 1.5)

Comparing point R with (k, 1.5), we get :

k = 2.

Hence, the value of k = 2.

Steps of construction :

1. Plot the points (1, -2) and (3, 4) on graph paper.

2. Connect the points through a straight line.

3. Take a point C (x = 2) and draw a line parallel to y-axis touching the graph at point B.

4. From point B draw a line parallel to x-axis and touching y-axis at point A (y = 1).

From graph,

B = (2, 1)

Comparing it with (2, b) we get :

b = 1.

Also, the graph touches y-axis at (0, -5).

Comparing it with (a, -5) we get :

a = 0.

Hence, a = 0 and b = 1.

Given,

Equation :

⇒ 3x - 2y = 4

⇒ 2y = 3x - 4

⇒ y = $\frac{3}{2}$x - 2 .........(1)

When, x = -2, y = $\frac{3}{2} \times -2 - 2$ = -3 - 2 = -5,

x = 0, y = $\frac{3}{2} \times 0 - 2$ = 0 - 2 = -2,

x = 2, y = $\frac{3}{2} \times 2 - 2$ = 3 - 2 = 1.

Tables of values for equation (1)

x	-2	0	2
y	-5	-2	1

Steps of construction :

1. Plot the points (-2, -5), (0, -2) and (2, 1) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 5x - 2y = 0

⇒ 2y = 5x

⇒ y = $\frac{5}{2}x$ ..............(2)

When, x = -2, y = $\frac{5}{2} \times -2$ = -5,

x = 0, y = $\frac{5}{2} \times 0$ = 0,

x = 2, y = $\frac{5}{2} \times 2$ = 5.

Tables of values for equation (2)

x	-2	0	2
y	-5	0	5

Steps of construction :

1. Plot the points (-2, -5), (0, 0) and (2, 5) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(-2, -5).

Hence, the solution of the given equations is x = -2, y = -5.

Given,

Equation :

⇒ 2x - 7y = 6

⇒ 2x = 6 + 7y

⇒ x = $\frac{6 + 7y}{2}$ ..........(1)

When y = 0, x = $\frac{6 + 7 \times 0}{2} = \frac{6}{2}$ = 3,

y = -1, x = $\frac{6 + 7 \times -1}{2} = \frac{6 - 7}{2}$ = -0.5,

y = -2, x = $\frac{6 + 7 \times -2}{2} = \frac{6 - 14}{2} = -\frac{8}{2}$ = -4.

Table of values for equation (1)

x	3	-0.5	-4
y	0	-1	-2

Steps of construction :

1. Plot the points (3, 0), $(-\frac{1}{2}, -1)$ and (-4, -2) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 5x - 8y = -4

⇒ 5x = 8y - 4

⇒ x = $\frac{8}{5}$y - $\frac{4}{5}$ ..............(2)

When, y = 0, x = $\frac{8}{5} \times 0 - \frac{4}{5} = 0 - \frac{4}{5} =$ -0.8,

y = 3, x = $\frac{8}{5} \times 3 - \frac{4}{5} = \frac{24}{5} - \frac{4}{5} = \frac{20}{5}$ = 4,

y = -2, x = $\frac{8}{5} \times -2 - \frac{4}{5} = -\frac{16}{5} - \frac{4}{5} = -\frac{20}{5}$ = -4.

Table of values for equation (2)

x	-0.8	4	-4
y	0	3	-2

Steps of construction :

1. Plot the points (-0.8, 0), (4, 3) and (-4, -2) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(-4, -2).

Hence, the solution of the given equations is x = -4, y = -2.

Given,

Equation :

⇒ x + y = 0

⇒ y = -x .............(1)

When, x = 0, y = 0,

x = 1, y = -(1) = -1,

x = 2, y = -(2) = -2.

Table of values for equation (1)

x	0	1	2
y	0	-1	-2

Steps of construction :

1. Plot the points (0, 0), (1, -1) and (2, -2) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 3x - 2y = 10

⇒ 2y = 3x - 10

⇒ y = $\frac{3}{2}x - 5$ ............(2)

When, x = 0, y = $\frac{3}{2} \times 0 - 5$ = 0 - 5 = -5,

x = 2, $y = \frac{3}{2} \times 2 - 5$ = 3 - 5 = -2,

x = 4, $y = \frac{3}{2} \times 4 - 5$ = 6 - 5 = 1.

Table of values for equation (2)

x	0	2	4
y	-5	-2	1

Steps of construction :

1. Plot the points (0, -5), (2, -2) and (4, 1) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(2, -2).

Hence, the solution of the given equations is x = 2, y = -2.

Given,

Equation :

⇒ 4x - 5y = -4

⇒ 5y = 4x + 4

⇒ y = $\frac{4x + 4}{5}$.............(1)

When, x = -1, y = $\frac{4 \times -1 + 4}{5} = \frac{0}{5}$ = 0,

x = 1.5, y = $\frac{4 \times 1.5 + 4}{5} = \frac{10}{5}$ = 2,

x = 4, y = $\frac{4 \times 4 + 4}{5} =\frac{20}{5}$ = 4.

Table of values for equation (1)

x	-1	1.5	4
y	0	2	4

Steps of construction :

1. Plot the points (-1, 0), (1.5, 2) and (4, 4) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 3x = 2y - 3

⇒ 2y = 3x + 3

⇒ y = $\frac{3x + 3}{2}$ ............(2)

When, x = -1, y = $\frac{3 \times -1 + 3}{2} = \frac{-3 + 3}{2} = \frac{0}{2}$ = 0,

x = 0, $y = \frac{3 \times 0 + 3}{2} = \frac{3}{2}$ = 1.5,

x = 1, $y = \frac{3 \times 1 + 3}{2} = \frac{6}{2}$ = 3.

Table of values for equation (2)

x	-1	0	1
y	0	1.5	3

Steps of construction :

1. Plot the points (-1, 0), (0, 1.5) and (1, 3) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(-1, 0).

Hence, the solution of the given equations is x = -1, y = 0.

Given,

Equation :

x + 3y = 8

x = 8 - 3y ..........(1)

When, y = 1, x = 8 - 3(1) = 8 - 3 = 5,

y = 2, x = 8 - 3(2) = 8 - 6 = 2,

y = 3, x = 8 - 3(3) = 8 - 9 = -1.

Table of values for equation (1)

x	-1	2	5
y	3	2	1

Steps of construction :

1. Plot the points (-1, 3), (2, 2) and (5, 1) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 3x = 2 + 2y

⇒ x = $\frac{2 + 2y}{3}$ ..........(2)

When, y = -1, x = $\frac{2 + 2 \times -1}{3} = \frac{2 - 2}{3}$ = 0.

y = 2, x = $\frac{2 + 2 \times 2}{3} = \frac{6}{3}$ = 2,

y = 5, x = $\frac{2 + 2 \times 5}{3} = \frac{12}{3}$ = 4.

Table of values for equation (2)

x	0	2	4
y	-1	2	5

Steps of construction :

1. Plot the points (0, -1), (2, 2) and (4, 5) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(2, 2).

Hence, the solution of the given equations is x = 2, y = 2..

Given,

⇒ 3y = 5 - x

⇒ y = $\frac{5 - x}{3}$ .......(1)

When x = -4, y = $\frac{5 - (-4)}{3} = \frac{9}{3}$ = 3,

x = -1, y = $\frac{5 - (-1)}{3} = \frac{6}{3}$ = 2,

x = 2, y = $\frac{5 - 2}{3} = \frac{3}{3}$ = 1.

Table of values for equation (1)

x	-4	-1	2
y	3	2	1

Steps of construction :

1. Plot the points (-4, 3), (-1, 2) and (2, 1) on graph paper.

2. Connect points by straight line.

Given,

⇒ 2x = y + 3

⇒ y = 2x - 3 ...........(2)

When x = 0, y = 2 × 0 - 3 = 0 - 3 = -3,

x = 1, y = 2 × 1 - 3 = 2 - 3 = -1,

x = 2, y = 2 × 2 - 3 = 4 - 3 = 1.

Table of values for equation (2)

x	0	1	2
y	-3	-1	1

Steps of construction :

1. Plot the points (0, -3), (1, -1) and (2, 1) on graph paper.

2. Connect points by straight line.

From graph,

The two lines meet at point P(2, 1).

Hence, the solution of the given equations is x = 2, y = 1.

(i) Given,

⇒ x + y + 3 = 0

⇒ y = -(3 + x) .........(1)

When x = -1, y = -[3 + (-1)] = -(3 - 1) = -2,

x = 0, y = -(3 + 0) = -3,

x = 1, y = -(3 + 1) = -4.

Table of values for equation (1)

x	-1	0	1
y	-2	-3	-4

Steps of construction :

1. Plot the points (-1, -2), (0, -3) and (1, -4) on graph paper.

2. Connect points by straight line.

Given,

⇒ 3x - 2y + 4 = 0

⇒ 2y = 3x + 4

⇒ y = $\frac{3x + 4}{2}$

When x = -2, y = $\frac{3 \times -2 + 4}{2} = \frac{-6 + 4}{2} = \frac{-2}{2}$ = -1,

x = 0, y = $\frac{3 \times 0 + 4}{2} = \frac{4}{2}$ = 2,

x = 2, y = $\frac{3 \times 2 + 4}{2} = \frac{6 + 4}{2} = \frac{10}{2}$ = 5.

Table of values for equation (2)

x	-2	0	2
y	-1	2	5

Steps of construction :

1. Plot the points (-2, -1), (0, 2) and (2, 5) on graph paper.

2. Connect points by straight line.

(ii) From graph,

P(-2, -1) is the intersection of lines.

Hence, coordinates of point of intersection = (-2, -1).

(iii) From P, draw a perpendicular to y-axis.

As, 1 unit = 2 cm.

So, PQ = 2 unit = 4 cm, OQ = 1 unit = 2 cm.

In right angle triangle,

By pythagoras theorem,

⇒ OP² = PQ² + OQ²

⇒ OP² = 4² + 2²

⇒ OP² = 16 + 4

⇒ OP² = 20

⇒ OP = $\sqrt{20}$ = 4.5 cm

Hence, distance of the point of intersection of the lines from the origin = 4.5 cm.

Considering,

⇒ 2x - 3y + 2 = 4x + 1

⇒ 3y = 2x - 4x + 2 - 1

⇒ 3y = -2x + 1

⇒ y = $\frac{1 - 2x}{3}$ .........(1)

When, x = -4, y = $\frac{1 - 2 \times (-4)}{3} = \frac{1 + 8}{3} = \frac{9}{3}$ = 3,

x = -1, y = $\frac{1 - 2 \times (-1)}{3} = \frac{1 + 2}{3} = \frac{3}{3}$ = 1,

x = 2, y = $\frac{1 - 2 \times 2}{3} = \frac{1 - 4}{3} = \frac{-3}{3}$ = -1.

Table of values for equation (1)

x	-4	-1	2
y	3	1	-1

Steps of construction :

1. Plot the points (-4, 3), (-1, 1) and (2, -1) on graph paper.

2. Connect points by straight line.

Considering,

⇒ 4x + 1 = 3x - y + 2

⇒ y = 3x - 4x + 2 - 1

⇒ y = -x + 1

⇒ y = 1 - x ............(2)

When, x = 0, y = 1 - 0 = 1,

x = 1, y = 1 - 1 = 0,

x = 2, y = 1 - 2 = -1.

Table of values for equation (2)

x	0	1	2
y	1	0	-1

Steps of construction :

1. Plot the points (0, 1), (1, 0) and (2, -1) on graph paper.

2. Connect points by straight line.

From graph,

The two lines intersect at P(2, -1).

Hence, the solution of the given equations is x = 2, y = -1.

(i) Given,

⇒ 3x - y - 2 = 0

⇒ y = 3x - 2 ............(1)

When x = 0, y = 3 × 0 - 2 = 0 - 2 = -2,

x = 1, y = 3 × 1 - 2 = 3 - 2 = 1,

x = 2, y = 3 × 2 - 2 = 6 - 2 = 4.

Table of values for equation (1)

x	0	1	2
y	-2	1	4

Steps of construction :

1. Plot the points (0, -2), (1, 1) and (2, 4).

2. Join the points.

Given,

⇒ 2x + y - 8 = 0

⇒ y = 8 - 2x ...........(2)

When x = 2, y = 8 - 2 × 2 = 8 - 4 = 4,

x = 3, y = 8 - 2 × 3 = 8 - 6 = 2,

x = 4, y = 8 - 2 × 4 = 8 - 8 = 0.

Table of values for equation (2)

x	2	3	4
y	4	2	0

Steps of construction :

1. Plot the points (2, 4), (3, 2) and (4, 0).

2. Join the points.

(ii) From the graph,

P(2, 4) is the point of intersection of lines.

BC = $3\frac{2}{3}$ cm

PV = 4 cm.

Area of triangle = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × PV

= $\frac{1}{2} \times 3\frac{2}{3}$ × 4

= $6\frac{2}{3}$ cm².

Hence, P(2, 4) is the point of intersection of lines and area = $6\frac{2}{3}$ cm².

Given,

⇒ 2x - y - 4 = 0

⇒ y = 2x - 4 ............(1)

When x = 1, y = 2(1) - 4 = 2 - 4 = -2,

x = 2, y = 2(2) - 4 = 4 - 4 = 0,

x = 3, y = 2(3) - 4 = 6 - 4 = 2.

Table of values for equation (1)

x	1	2	3
y	-2	0	2

Steps of construction :

1. Plot the points (1, -2), (2, 0) and (3, 2).

2. Join the points.

Given,

⇒ x + y + 1 = 0

⇒ y = -(x + 1) .............(2)

When, x = -1, y = -(-1 + 1) = 0,

x = 0, y = -(0 + 1) = -1,

x = 1, y = -(1 + 1) = -2.

Table of values for equation (2)

x	-1	0	1
y	0	-1	-2

Steps of construction :

1. Plot the points (-1, 0), (0, -1) and (1, -2).

2. Join the points.

From graph,

A(1, -2) is the point of intersection of lines.

ABC are the vertices of triangle.

From A, draw AD perpendicular to BC.

AD = 1 unit and BC = 3 units.

Area of △ABC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AD

= $\frac{1}{2}$ × 3 × 1

= $\frac{3}{2}$ sq. units

Hence, point of intersection = (1, -2) and area of triangle = $\frac{3}{2}$ sq. units

Given,

⇒ x + 2y = 4

⇒ 2y = 4 - x

⇒ y = $\frac{4 - x}{2}$ ......................(1)

When x = 0, y = $\frac{4 - 0}{2} = \frac{4}{2}$ = 2,

x = 2, y = $\frac{4 - 2}{2} = \frac{2}{2}$ = 1,

x = 4, y = $\frac{4 - 4}{2} = \frac{0}{2}$ = 0.

Table of values for equation (1)

x	0	2	4
y	2	1	0

Steps of construction :

1. Plot the points (0, 2), (2, 1) and (4, 0).

2. Join the points.

Given,

⇒ 3x - 2y = 4

⇒ 2y = 3x - 4

⇒ y = $\frac{3x - 4}{2}$ ..................(2)

When x = 0, y = $\frac{3 \times 0 - 4}{2} = \frac{0 - 4}{2} = \frac{-4}{2}$ = -2,

x = 2, y = $\frac{3 \times 2 - 4}{2} = \frac{6 - 4}{2} = \frac{2}{2}$ = 1,

x = 4, y = $\frac{3 \times 4 - 4}{2} = \frac{12 - 4}{2} = \frac{8}{2}$ = 4.

x = $\frac{4}{3}, y = \frac{3 \times \frac{4}{3} - 4}{2} = \frac{4 - 4}{2} = \frac{0}{2}$ = 0.

Table of values for equation (2)

x	0	2	4	$\frac{4}{3}$
y	-2	1	4	0

Steps of construction :

1. Plot the points (0, -2), (2, 1), (4, 4) and $\Big(\frac{4}{3}, 0\Big)$.

2. Join the points.

From graph,

A(2, 1) is the point of intersection of lines.

AEF is the triangle between lines and y-axis.

From A, draw AG perpendicular to EF.

Using distance formula, distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance between E (0, 2) and F (0, -2)

= $\sqrt{(0 - 0)^2 + \Big(2 - (-2)\Big)^2} = \sqrt{(2 + 2)^2} = \sqrt{4^2} = 4$ units

From graph,

AG = 2 units

EF = 4 units

Area of triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x EF x AG

= $\frac{1}{2}$ x 4 x 2

= 1 x 4

= 4 sq. units.

And, ABC is the triangle between lines and x-axis.

Using distance formula, distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance between B $\Big(\frac{4}{3}, 0\Big)$ and C (4, 0)

= $\sqrt{\Big(4 - \frac{4}{3}\Big)^2 + (0 - 0)^2} = \sqrt{\Big(\frac{12 - 4}{3}\Big)^2} = \sqrt{\Big(\frac{8}{3}\Big)^2} = \frac{8}{3}$

From A, draw AD perpendicular to BC.

AD = 1 unit

BC = $\frac{8}{3}$ units

Area of triangle = $\frac{1}{2}$ x base x height

$$\begin{aligned}\text{Area of triangle ABC} = \frac{1}{2} \times BC \times AD \\[1em] = \frac{1}{2} \times \frac{8}{3} \times 1 \\[1em] = \frac{4}{3} \text{sq. units}\end{aligned}$$

Hence, point of intersection of lines is x = 2, y = 1 and area of the triangle formed by the lines and the y-axis = 4 sq. unit and area of the triangle formed by the lines and the x-axis = $\frac{4}{3}$ sq units.

Given,

1st equation :

⇒ x + 3 = 0

⇒ x = -3 ............(1)

2nd equation :

⇒ y - 2 = 0

⇒ y = 2 .............(2)

3rd equation :

⇒ 2x + 3y = 12

⇒ 3y = 12 - 2x

⇒ y = $\frac{12 - 2x}{3}$ .........(3)

When, x = -3, y = $\frac{12 - 2 \times (-3)}{3} = \frac{12 + 6}{3} = \frac{18}{3}$ = 6,

x = 0, y = $\frac{12 - 2 \times 0}{3} = \frac{12}{3}$ = 4,

x = 3, y = $\frac{12 - 2 \times 3}{3} = \frac{12 - 6}{3} = \frac{6}{3}$ = 2.

Table of values of equation (3) :

x	-3	0	3
y	6	4	2

Steps of construction :

1. Plot the points (-3, 6), (0, 4) and (3, 2).

2. Join the points.

From graph,

ABC is the required triangle.

Hence, coordinates of vertices of triangle formed by these lines = (-3, 6), (-3, 2) and (3, 2).

Given,

1st equation :

y = 0,

2nd equation :

y = x,

When, x = 0, y = 0,

x = 1, y = 1,

x = 2, y = 2.

Table of values of equation (2) :

x	0	1	2
y	0	1	2

Steps of construction :

1. Plot the points (0, 0), (1, 1) and (2, 2).

2. Join the points.

3rd equation :

⇒ 2x + 3y = 10

⇒ 3y = 10 - 2x

⇒ y = $\frac{10 - 2x}{3}$ ..........(3)

When, x = -1, y = $\frac{10 - 2 \times (-1)}{3} = \frac{10 + 2}{3} = \frac{12}{3}$ = 4,

x = 2, y = $\frac{10 - 2 \times 2}{3} = \frac{10 - 4}{3} = \frac{6}{3}$ = 2,

x = 5, y = $\frac{10 - 2 \times 5}{3} = \frac{10 - 10}{3}$ = 0.

Table of values of equation (3) :

x	-1	2	5
y	4	2	0

Steps of construction :

1. Plot the points (-1, 4), (2, 2) and (5, 0).

2. Join the points.

From graph,

ABC is the triangle.

From A, draw AD perpendicular to x-axis.

AD = 2 units and BC = 5 units.

Area of triangle = $\frac{1}{2}$ × base × height

= $\frac{1}{2} \times BC \times AD$

= $\frac{1}{2} \times 5 \times 2$

= 5 sq. units.

Hence, coordinates of triangle are (0, 0), (5, 0) and (2, 2) and area = 5 sq. units.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (2, 3)

Co-ordinates of Q = (4, 1)

$$\begin{aligned}PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2} \\[1em] = \sqrt{2^2 + (-2)^2} \\[1em] = \sqrt{4 + 4}\\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.\end{aligned}$$

Hence, distance between (2, 3) and (4, 1) = $2\sqrt{2}$ units.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (0, 0)

Co-ordinates of Q = (36, 15)

$$\begin{aligned}PQ = \sqrt{(36 - 0)^2 + (15 - 0)^2} \\[1em] = \sqrt{36^2 + 15^2} \\[1em] = \sqrt{1296 + 225}\\[1em] = \sqrt{1521} \\[1em] = 39.\end{aligned}$$

Hence, distance between (0, 0) and (36, 15) = 39 units.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (a, b)

Co-ordinates of Q = (-a, -b)

$$\begin{aligned}PQ = \sqrt{(-a - a)^2 + (-b - b)^2} \\[1em] = \sqrt{(-2a)^2 + (-2b)^2} \\[1em] = \sqrt{4a^2 + 4b^2}\\[1em] = \sqrt{4(a^2 + b^2)} \\[1em] = 2\sqrt{a^2 + b^2}.\end{aligned}$$

Hence, distance between (a, b) and (-a, -b) = $2\sqrt{a^2 + b^2}$ units.

Since, point A is on y-axis so its x-coordinate = 0.

Given,

Ordinate of A = 4.

∴ A = (0, 4).

Since, point B is on x-axis so its y-coordinate = 0.

Given,

Abscissa of B = -3.

∴ B = (-3, 0).

By distance formula,

$$\begin{aligned}\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(-3 - 0)^2 + (0 - 4)^2} \\[1em] = \sqrt{(-3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.\end{aligned}$$

Hence, AB = 5 units.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}\Rightarrow AB = \sqrt{[a - (-3)]^2 + [(-5) - (-14)]^2} \\[1em] \Rightarrow 9 = \sqrt{(a + 3)^2 + (-5 + 14)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 9 + 6a + (9)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 6a + 9 + 81}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow a^2 + 6a + 90 = (9)^2 \\[1em] \Rightarrow a^2 + 6a + 90 = 81 \\[1em] \Rightarrow a^2 + 6a + 90 - 81 = 0 \\[1em] \Rightarrow a^2 + 6a + 9 = 0 \\[1em] \Rightarrow a^2 + 3a + 3a + 9 = 0 \\[1em] \Rightarrow a(a + 3) + 3(a + 3) = 0 \\[1em] \Rightarrow (a + 3)(a + 3) = 0 \\[1em] \Rightarrow a + 3 = 0 \\[1em] \Rightarrow a = -3.\end{aligned}$$

Hence, value of a = -3.

We know that,

y-coordinate of any point on x-axis = 0.

Let point on x-axis which is at a distance of 5 units from (5, -4) be P(x, 0).

By distance formula,

$$\begin{aligned}\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 5 = \sqrt{(x - 5)^2 + [0 - (-4)]^2} \\[1em] \Rightarrow 5 = \sqrt{x^2 + 25 - 10x + 16} \\[1em] \Rightarrow (5)^2 = x^2 - 10x + 41\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 25 = x^2 - 10x + 41 \\[1em] \Rightarrow x^2 - 10x + 41 - 25 = 0 \\[1em] \Rightarrow x^2 - 10x + 16 = 0 \\[1em] \Rightarrow x^2 - 8x - 2x + 16 = 0 \\[1em] \Rightarrow x(x - 8) - 2(x - 8) = 0 \\[1em] \Rightarrow (x - 2)(x - 8) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x - 8 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = 8.\end{aligned}$$

∴ P = (x, 0) = (2, 0) or (8, 0).

Hence, points on the x-axis which are at a distance of 5 units from the point (5, -4) are (2, 0) or (8, 0).

We know that,

x-coordinate of any point on y-axis = 0.

Let point on y-axis which is at a distance of 10 units from (8, 8) be P(0, y).

By distance formula,

$$\begin{aligned}\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 10 = \sqrt{(0 - 8)^2 + (y - 8)^2} \\[1em] \Rightarrow 10 = \sqrt{(-8)^2 + y^2 + 8^2 - 16y} \\[1em] \Rightarrow 10 = \sqrt{64 + y^2 + 64 - 16y} \\[1em]\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 10^2 = 64 + y^2 + 64 - 16y \\[1em] \Rightarrow 100 = y^2 + 128 - 16y \\[1em] \Rightarrow y^2 - 16y + 128 - 100 = 0 \\[1em] \Rightarrow y^2 - 16y + 28 = 0 \\[1em] \Rightarrow y^2 - 14y - 2y + 28 = 0 \\[1em] \Rightarrow y(y - 14) - 2(y - 14) = 0 \\[1em] \Rightarrow (y - 2)(y - 14) = 0 \\[1em] \Rightarrow y - 2 = 0 \text{ or } y - 14 = 0 \\[1em] \Rightarrow y = 2 \text{ or } y = 14.\end{aligned}$$

∴ P = (0, y) = (0, 2) or (0, 14).

Hence, points on the y-axis which are at a distance of 10 units from the point (8, 8) are (0, 2) or (0, 14).

Given,

Ordinate of the point is twice the abscissa.

Let abscissa of point be k, then ordinate = 2k

Let the required point be P.

∴ P = (k, 2k).

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

Distance between (4, 3) and (k, 2k) = $\sqrt{10}$ units.

$$\begin{aligned}\Rightarrow \sqrt{10} = \sqrt{(k - 4)^2 + (2k - 3)^2} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + (4)^2 - 8k + (2k)^2 + (3)^2 - 12k} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + 16 - 8k + 4k^2 + 9 - 12k}\end{aligned}$$

On squaring both sides

$$\begin{aligned}\Rightarrow 10 = k^2 + 16 - 8k + 4k^2 + 9 - 12k \Rightarrow 10 = 5k^2 - 20k + 25 \\[1em] \Rightarrow 10 = 5(k^2 - 4k + 5) \\[1em] \Rightarrow k^2 - 4k + 5 = 2 \\[1em] \Rightarrow k^2 - 4k + 5 - 2 = 0 \\[1em] \Rightarrow k^2 - 4k + 3 = 0 \\[1em] \Rightarrow k^2 - 3k - k + 3 = 0 \\[1em] \Rightarrow k(k - 3) - 1(k - 3) = 0 \\[1em] \Rightarrow (k - 1)(k - 3) = 0 \\[1em] \Rightarrow k - 1 = 0 \text{ or } k - 3 = 0 \\[1em] \Rightarrow k = 1 \text{ or } k = 3.\end{aligned}$$

When k = 1,

P = (k, 2k) = (1, 2).

When k = 3,

P = (k, 2k) = (3, 6).

Hence, required points are (1, 2) or (3, 6).

We know that,

y-coordinate of any point on x-axis = 0.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let point on x-axis which is equidistant from the points (2, -5) and (-2, 9) be P(x, 0).

∴ Distance between (2, -5) and (x, 0) = Distance between (-2, 9) and (x, 0).

$$\begin{aligned}\Rightarrow \sqrt{(x - 2)^2 + [0 - (-5)]^2} = \sqrt{[x - (-2)]^2 + (0 - 9)^2} \\[1em] \Rightarrow \sqrt{x^2 + 4 - 4x + 5^2} = \sqrt{(x + 2)^2 + (-9)^2} \\[1em] \Rightarrow \sqrt{x^2 - 4x + 25 + 4} = \sqrt{x^2 + 4 + 4x + 81} \\[1em]\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow x^2 - 4x + 29 = x^2 + 4x + 85 \\[1em] \Rightarrow x^2 - x^2 + 4x + 4x = 29 - 85 \\[1em] \Rightarrow 8x = -56 \\[1em] \Rightarrow x = -7.\end{aligned}$$

P = (x, 0) = (-7, 0).

Hence, required point is (-7, 0).

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given, PQ = QR.

∴ Distance between (6, -1) and (1, 3) = Distance between (1, 3) and (x, 8).

$$\begin{aligned}\Rightarrow \sqrt{(1 - 6)^2 + [3 - (-1)]^2} = \sqrt{(x - 1)^2 + (8 - 3)^2} \\[1em] \Rightarrow \sqrt{(-5)^2 + [4]^2} = \sqrt{x^2 + 1 - 2x + (5)^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 - 2x + 1 + 25} \\[1em] \Rightarrow \sqrt{41} = \sqrt{x^2 - 2x + 26}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 41 = x^2 - 2x + 26 \\[1em] \Rightarrow x^2 - 2x + 26 - 41 = 0 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3(x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0 \\[1em] \Rightarrow x = -3 \text{ or } x = 5.\end{aligned}$$

Hence, x = -3 or 5.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

∴ Distance between Q(0, 1) and P(5, -3) = Distance between Q(0, 1) and R(x, 6).

$$\begin{aligned}\Rightarrow \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{(x - 0)^2 + (6 - 1)^2} \\[1em] \Rightarrow \sqrt{5^2 + (-4)^2} = \sqrt{x^2 + 5^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 + 25}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 25 + 16 = x^2 + 25 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x = \sqrt{16} \\[1em] \Rightarrow x = \pm 4.\end{aligned}$$

Hence, x = ±4.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

∴ Distance between (x, y) and (7, 1) = Distance between (x, y) and (3, 5).

$$\begin{aligned}\therefore \sqrt{(7 - x)^2 + (1 - y)^2} = \sqrt{(3 - x)^2 + (5 - y)^2} \\[1em] \Rightarrow \sqrt{49 + x^2 - 14x + 1 + y^2 - 2y} = \sqrt{9 + x^2 - 6x + 25 + y^2 - 10y}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 49 + x^2 - 14x + 1 + y^2 - 2y = 9 + x^2 - 6x + 25 + y^2 - 10y \\[1em] \Rightarrow x^2 + y^2 - 14x - 2y + 50 = x^2 + y^2 - 6x - 10y + 34 \\[1em] \Rightarrow x^2 + y^2 - x^2 - y^2 - 6x + 14x = 10y - 2y + 50 - 34 \\[1em] \Rightarrow 8x = 8y + 16 \\[1em] \Rightarrow 8x = 8(y + 2) \\[1em] \Rightarrow x = y + 2 \\[1em] \Rightarrow x - y = 2.\end{aligned}$$

Hence, relation between x and y is x - y = 2.

Let y-coordinate of point P be k.

∴ x-coordinate = 2k.

P = (2k, k)

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

P is equidistant from the points Q(2, -5) and R(-3, 6).

$$\begin{aligned}\Rightarrow \sqrt{(2 - 2k)^2 + (-5 - k)^2} = \sqrt{(-3 - 2k)^2 + (6 - k)^2} \\[1em] \Rightarrow \sqrt{4 + 4k^2 - 8k + 25 + k^2 + 10k} = \sqrt{9 + 4k^2 + 12k + 36 + k^2 - 12k} \\[1em] \Rightarrow \sqrt{5k^2 + 2k + 29} = \sqrt{5k^2 + 45}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 5k^2 + 2k + 29 = 5k^2 + 45 \\[1em] \Rightarrow 5k^2 - 5k^2 + 2k = 45 - 29 \\[1em] \Rightarrow 2k = 16 \\[1em] \Rightarrow k = \frac{16}{2} \\[1em] \Rightarrow k = 8.\end{aligned}$$

P = (2k, k) = (16, 8).

Hence, coordinates of P = (16, 8).

Given,

A(4, 3) and B(x, 5) are on a circle with center C(2, 3).

∴ AC = BC [Radius of same circle]

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting values in above equation, we get :

$$\begin{aligned}\Rightarrow \sqrt{(2 - 4)^2 + (3 - 3)^2} = \sqrt{(2 - x)^2 + (3 - 5)^2} \\[1em] \Rightarrow \sqrt{(-2)^2} = \sqrt{4 + x^2 - 4x + (-2)^2} \\[1em] \Rightarrow \sqrt{4} = \sqrt{4 + x^2 - 4x + 4}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 4 = x^2 - 4x + 8 \\[1em] \Rightarrow x^2 - 4x + 4 = 0 \\[1em] \Rightarrow x^2 - 2x - 2x + 4 = 0 \\[1em] \Rightarrow x(x - 2) - 2(x - 2) = 0 \\[1em] \Rightarrow (x - 2)(x - 2) = 0 \\[1em] \Rightarrow x - 2 = 0 \\[1em] \Rightarrow x = 2.\end{aligned}$$

Hence, value of x = 2.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

AB = AC.

$$\begin{aligned}\Rightarrow \sqrt{(3 - 0)^2 + (p - 2)^2} = \sqrt{(p - 0)^2 + (5 - 2)^2} \\[1em] \Rightarrow \sqrt{3^2 + p^2 + 4 - 4p} = \sqrt{p^2 + 3^2} \\[1em] \Rightarrow \sqrt{9 + p^2 + 4 - 4p} = \sqrt{p^2 + 3^2}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow p^2 + 13 - 4p = p^2 + 9 \\[1em] \Rightarrow p^2 - p^2 + 4p = 13 - 9 \\[1em] \Rightarrow 4p = 4 \\[1em] \Rightarrow p = 1.\end{aligned}$$

Hence, p = 1.

Given,

Center (O) = (3, 3)

Let, A = (6, 2), B = (0, 4) and C = (4, 6).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow AO = \sqrt{(6 - 3)^2 + (2 - 3)^2} \\[1em] = \sqrt{(3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow BO = \sqrt{(0 - 3)^2 + (4 - 3)^2} \\[1em] = \sqrt{(-3)^2 + 1^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow CO = \sqrt{(4 - 3)^2 + (6 - 3)^2} \\[1em] = \sqrt{1^2 + 3^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10}. \\[1em]\end{aligned}$$

Since, AO = BO = CO.

Hence, proved that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).

Given,

Diameter = 20 units

Radius = $\frac{\text{Diameter}}{2} = \frac{20}{2}$ = 10 units.

∴ AC = 10 units

$$\begin{aligned}\Rightarrow \sqrt{[-3 - (2α - 1)]^2 + [-1 - (3α + 1)]^2} = 10 \\[1em] \Rightarrow \sqrt{(-3 - 2α + 1)^2 + (-1 - 3α - 1)^2} = 10 \\[1em] \Rightarrow \sqrt{(-2 - 2α)^2 + (-3α - 2)^2} = 10 \\[1em] \Rightarrow \sqrt{4 + 4α^2 + 8α + 9α^2 + 4 + 12α} = 10 \\[1em] \Rightarrow \sqrt{13α^2 + 20α + 8} = 10\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 13α^2 + 20α + 8 = 10^2 \\[1em] \Rightarrow 13α^2 + 20α + 8 = 100 \\[1em] \Rightarrow 13α^2 + 20α + 8 - 100 = 0 \\[1em] \Rightarrow 13α^2 + 20α - 92 = 0 \\[1em] \Rightarrow 13α^2 + 46α - 26α - 92 =0 \\[1em] \Rightarrow α(13α + 46) - 2(13α + 46) = 0 \\[1em] \Rightarrow (13α + 46)(α - 2) = 0 \\[1em] \Rightarrow 13α + 46 = 0 \text{ or } α - 2 = 0 \\[1em] \Rightarrow 13α = -46 \text{ or } α = 2 \\[1em] \Rightarrow α = -\frac{46}{13} \text{ or } α = 2.\end{aligned}$$

Hence, α = $-\frac{46}{13}$ or α = 2.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{(6 - 3)^2 + (4 - 1)^2} \\[1em] = \sqrt{3^2 + 3^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}. \\[1em] BC = \sqrt{(8 - 6)^2 + (6 - 4)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] AC = \sqrt{(8 - 3)^2 + (6 - 1)^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}. \\[1em] \Rightarrow AB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC.\end{aligned}$$

Thus, AC = AB + BC.

Hence, proved that A, B and C are collinear.

Let points be A(5, -2), B(6, 4) and C(7, -2).

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{(6 - 5)^2 + [4 - (-2)]^2} \\[1em] = \sqrt{1^2 + 6^2} \\[1em] = \sqrt{1 + 36} \\[1em] = \sqrt{37}. \\[1em] BC = \sqrt{(7 - 6)^2 + [-2 - 4]^2} \\[1em] = \sqrt{1^2 + (-6)^2} \\[1em] = \sqrt{37}.\end{aligned}$$

Since, AB = BC.

Hence, proved that (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{[-4 - (-5)]^2 + [-2 - 6]^2} \\[1em] = \sqrt{(-4 + 5)^2 + (-8)^2} \\[1em] = \sqrt{1^2 + 64} \\[1em] = \sqrt{65}. \\[1em] BC = \sqrt{[7 - (-4)]^2 + [5 - (-2)]^2} \\[1em] = \sqrt{11^2 + 7^2} \\[1em] = \sqrt{121 + 49} \\[1em] = \sqrt{170}. \\[1em] AC = \sqrt{[7 - (-5)]^2 + (5 - 6)^2} \\[1em] = \sqrt{12^2 + (-1)^2} \\[1em] = \sqrt{144 + 1} \\[1em] = \sqrt{145}.\end{aligned}$$

Since, AB ≠ BC ≠ AC.

Hence, ABC is a scalene triangle.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] BC = \sqrt{[-\sqrt{3} - (-1)]^2 + [\sqrt{3} - (-1)]^2} \\[1em] = \sqrt{[-\sqrt{3} + 1]^2 + [\sqrt{3} + 1]^2} \\[1em] = \sqrt{3 + 1 - 2\sqrt{3} + 3 + 1 + 2\sqrt{3}} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] AC = \sqrt{[-\sqrt{3} - 1]^2 + [\sqrt{3} - 1]^2} \\[1em] = \sqrt{3 + 1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3}} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em]\end{aligned}$$

Since, AB = BC = AC.

Hence, ABC is an equilateral triangle.

Let points be A(7, 10), B(-2, 5) and C(3, -4).

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{(-2 - 7)^2 + (5 - 10)^2} \\[1em] = \sqrt{(-9)^2 + (-5)^2} \\[1em] = \sqrt{81 + 25} \\[1em] = \sqrt{106} \text{ units}. \\[1em] BC = \sqrt{[3 - (-2)]^2 + [-4 - 5]^2} \\[1em] = \sqrt{5^2 + (-9)^2} \\[1em] = \sqrt{25 + 81} \\[1em] = \sqrt{106} \text{ units}. \\[1em] AC = \sqrt{(3 - 7)^2 + (-4 - 10)^2} \\[1em] = \sqrt{(-4)^2 + (-14)^2} \\[1em] = \sqrt{16 + 196} \\[1em] = \sqrt{212} \text{ units}.\end{aligned}$$

⇒ AB² + BC² = $(\sqrt{106})^2 + (\sqrt{106})^2$

= 106 + 106

= 212

= AC².

Since, AB = BC and AB² + BC² = AC².

∴ ABC is an isosceles right angle triangle, right angled at B.

Hence, proved that (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles triangle.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{(-2 - 0)^2 + (a - 3)^2} \\[1em] = \sqrt{(-2)^2 + a^2 + 9 - 6a} \\[1em] = \sqrt{4 + a^2 + 9 - 6a} \\[1em] = \sqrt{a^2 - 6a + 13}. \\[1em] BC = \sqrt{[-1 - (-2)]^2 + (4 - a)^2} \\[1em] = \sqrt{(-1 + 2)^2 + (4 - a)^2} \\[1em] = \sqrt{(1)^2 + 16 + a^2 - 8a} \\[1em] = \sqrt{1 + 16 + a^2 - 8a} \\[1em] = \sqrt{a^2 - 8a + 17}. \\[1em] AC = \sqrt{(-1 - 0)^2 + (4 - 3)^2} \\[1em] = \sqrt{(-1)^2 + (1)^2} \\[1em] = \sqrt{1 + 1} \\[1em] = \sqrt{2}.\end{aligned}$$

By pythagoras theorem,

AB² + AC² = BC²

$$\begin{aligned}\Rightarrow \Big(\sqrt{a^2 - 6a + 13}\Big)^2 + \Big(\sqrt{2}\Big)^2 = \Big(\sqrt{a^2 - 8a + 17}\Big)^2 \\[1em] \Rightarrow a^2 - 6a + 13 + 2 = a^2 - 8a + 17 \\[1em] \Rightarrow a^2 - a^2 - 6a + 8a = 17 - 15 \\[1em] \Rightarrow 2a = 2 \\[1em] \Rightarrow a = 1.\end{aligned}$$

Hence, value of a = 1.

Let A(0, -1), B(-2, 3), C(6, 7) and D(8, 3).

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{(-2 - 0)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{(-2)^2 + [3 + 1]^2}\\[1em] = \sqrt{4 + 4^2} \\[1em] = \sqrt{4 + 16} \\[1em] = \sqrt{20}. \\[1em] BC = \sqrt{[6 - (-2)]^2 + (7 - 3)^2} \\[1em] = \sqrt{[6 + 2]^2 + (4)^2} \\[1em] = \sqrt{8^2 + 4^2} \\[1em] = \sqrt{64 + 16} \\[1em] = \sqrt{80}. \\[1em] CD = \sqrt{(8 - 6)^2 + (3 - 7)^2} \\[1em] = \sqrt{2^2 + (-4)^2} \\[1em] = \sqrt{4 + 16} \\[1em] = \sqrt{20} \\[1em] AD = \sqrt{(8 - 0)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{8^2 + [3 + 1]^2} \\[1em] = \sqrt{64 + 4^2} \\[1em] = \sqrt{64 + 16} \\[1em] = \sqrt{80}.\end{aligned}$$

Since, AB = CD and BC = AD.

∴ ABCD is a rectangle.

Area of rectangle ABCD = AB × BC

$$\begin{aligned}= \sqrt{20} \times \sqrt{80} \\[1em] = 2\sqrt{5} \times 4\sqrt{5} \\[1em] = 8 \times 5 \\[1em] = 40 \text{ sq. units}.\end{aligned}$$

Hence, area of rectangle = 40 sq. units.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}PQ = \sqrt{(3 - 2)^2 + [4 - (-1)]^2} \\[1em] = \sqrt{1^2 + [4 + 1]^2} \\[1em] = \sqrt{1 + 5^2} \\[1em] = \sqrt{1 + 25} \\[1em] = \sqrt{26}. \\[1em] QR = \sqrt{(-2 - 3)^2 + (3 - 4)^2} \\[1em] = \sqrt{(-5)^2 + (-1)^2} \\[1em] = \sqrt{25 + 1} \\[1em] = \sqrt{26}. \\[1em] RS = \sqrt{[-3 - (-2)]^2 + [-2 - 3]^2} \\[1em] = \sqrt{[-3 + 2]^2 + [-5]^2} \\[1em] = \sqrt{[-1]^2 + 25} \\[1em] = \sqrt{26}. \\[1em] PS = \sqrt{[-3 - 2]^2 + [-2 - (-1)]^2} \\[1em] = \sqrt{[-5]^2 + [-2 + 1]^2} \\[1em] = \sqrt{25 + [-1]^2} \\[1em] = \sqrt{25 + 1} \\[1em] = \sqrt{26}.\end{aligned}$$

Here, PQ = QR = RS = PS.

So, PQRS can be a square or rhombus.

Calculating diagonal,

$$\begin{aligned}PR = \sqrt{(-2 - 2)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{(-4)^2 + [3 + 1]^2} \\[1em] = \sqrt{16 + 4^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = 4\sqrt{2}. \\[1em] QS = \sqrt{(-3 - 3)^2 + (-2 - 4)^2} \\[1em] = \sqrt{(-6)^2 + (-6)^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} \\[1em] = 6\sqrt{2}.\end{aligned}$$

PR ≠ QS.

Since, diagonal of rhombus are not equal,

∴ PQRS is a rhombus

Area of rhombus = $\frac{1}{2} \times d_1 \times d_2$

= $\frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2}$

= $\frac{48}{2}$ = 24 sq. units.

Hence, PQRS is a rhombus and area = 24 sq. units.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}AB = \sqrt{(-2 - 2)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-4)^2 + (-1)^2} \\[1em] = \sqrt{16 + 1} \\[1em] = \sqrt{17}. \\[1em] BC = \sqrt{[-1 - (-2)]^2 + [-2 - 2]^2} \\[1em] = \sqrt{[-1 + 2]^2 + [-4]^2} \\[1em] = \sqrt{1^2 + 16} \\[1em] = \sqrt{17}. \\[1em] CD = \sqrt{[3 - (-1)]^2 + [-1 - (-2)]^2} \\[1em] = \sqrt{[3 + 1]^2 + [-1 + 2]^2} \\[1em] = \sqrt{4^2 + 1^2} \\[1em] = \sqrt{16 + 1} \\[1em] = \sqrt{17}. \\[1em] AD = \sqrt{(3 - 2)^2 + (-1 - 3)^2} \\[1em] = \sqrt{1^2 + (-4)^2} \\[1em] = \sqrt{1 + 16} \\[1em] = \sqrt{17}.\end{aligned}$$

Calculating diagonals,

$$\begin{aligned}AC = \sqrt{(-1 - 2)^2 + (-2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-5)^2} \\[1em] = \sqrt{9 + 25} \\[1em] = \sqrt{34}. \\[1em] BD = \sqrt{[3 - (-2)]^2 + (-1 - 2)^2} \\[1em] = \sqrt{[3 + 2]^2 + (-3)^2} \\[1em] = \sqrt{5^2 + 9} \\[1em] = \sqrt{25 + 9} \\[1em] = \sqrt{34}.\end{aligned}$$

Since, AB = BC =CD = AD and AC = BD.

Hence, proved that A, B, C and D are the vertices of a square.

Let coordinates be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0).

$$\begin{aligned}AB = \sqrt{[1 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[1 + 1]^2 + [0 + 2]^2}\\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] CD = \sqrt{[-3 - (-1)]^2 + [0 - 2]^2} \\[1em] = \sqrt{[-3 + 1]^2 + [-2]^2} \\[1em] = \sqrt{[-2]^2 + [-2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] AD = \sqrt{[-3 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[-3 + 1]^2 + [2]^2} \\[1em] =\sqrt{[-2]^2 + [2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}.\end{aligned}$$

Since, AB = BC = CD = AD.

So, ABCD can be square or rhombus.

Calculating diagonal,

$$\begin{aligned}AC = \sqrt{[-1 - (-1)]^2 + [2 - (-2)]^2} \\[1em] = \sqrt{[-1 + 1]^2 + [2 + 2]^2} \\[1em] = \sqrt{0 + 4^2} \\[1em] = \sqrt{16} \\[1em] = 4. \\[1em] BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 0^2} \\[1em] = \sqrt{16} \\[1em] = 4.\end{aligned}$$

Since, diagonals are also equal.

Hence, (-1, -2), (1, 0), (-1, 2), (-3, 0) are the vertices of a square.

Let A(4, 5), B(7, 6), C(4, 3), D(1, 2) are the given points.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} \\[1em] = \sqrt{3^2 + 1^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units}. \\[1em] BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units}. \\[1em] CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units}. \\[1em] AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} \\[1em] = \sqrt{3^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units}.\end{aligned}$$

Calculating diagonals :

$$\begin{aligned}AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} \\[1em] = \sqrt{0 + (-2)^2} \\[1em] = \sqrt{4} \\[1em] = 2 \text{ units}. \\[1em] BD = \sqrt{(1 - 7)^2 + (2 - 6)^2}\\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \text{ units}.\end{aligned}$$

Since, opposite sides are equal and diagonals are not equal.

Hence, (4, 5), (7, 6), (4, 3), (1, 2) are the vertices of a parallelogram.

Let O(x, y) be the circumcentre of the circle.

Let A(8, 6), B(8, -2) and C(2, -2) be the vertices of the triangle.

OB = OC [Radii of same circle]

By distance formula,

$$\begin{aligned}\Rightarrow \sqrt{(x - 8)^2 + [y - (-2)]^2} = \sqrt{(x - 2)^2 + [y - (-2)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 64 - 16x + [y + 2]^2} = \sqrt{x^2 + 4 - 4x + [y + 2]^2}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow x^2 + 64 - 16x + [y + 2]^2 = x^2 + 4 - 4x + [y + 2]^2 \\[1em] \Rightarrow x^2 - x^2 + [y + 2]^2 - [y + 2]^2 + 64 - 4 = 16x - 4x \\[1em] \Rightarrow 12x = 60 \\[1em] \Rightarrow x = \frac{60}{12} \\[1em] \Rightarrow x = 5.\end{aligned}$$

Also,

OA = OB [Radii of same circle]

By distance formula,

$$\begin{aligned}\Rightarrow \sqrt{(x - 8)^2 + (y - 6)^2} = \sqrt{(x - 8)^2 + [y - (-2)]^2} \\[1em] \Rightarrow \sqrt{(x - 8)^2 + y^2 + 36 - 12y} = \sqrt{(x - 8)^2 + [y + 2]^2}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow (x - 8)^2 + y^2 + 36 - 12y = (x - 8)^2 + [y + 2]^2 \\[1em] \Rightarrow (x - 8)^2 + y^2 + 36 - 12y = (x - 8)^2 + y^2 + 4 + 4y \\[1em] \Rightarrow (x - 8)^2 - (x - 8)^2 + y^2 - y^2 + 4y + 12y = 36 - 4 \\[1em] \Rightarrow 16y = 32 \\[1em] \Rightarrow y = \frac{32}{16} \\[1em] \Rightarrow y = 2.\end{aligned}$$

O = (x, y) = (5, 2).

Radius = OA.

$$\begin{aligned}OA = \sqrt{(8 - 5)^2 + (6 - 2)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.\end{aligned}$$

Hence, circumcenter = (5, 2) and circumradius = 5 units.

In second quadrant,

x-coordinate is negative and y-coordinate is positive.

∴ P(-3, 5) lies in second quadrant.

Hence, Option 2 is the correct option.

x-coordinate of any point on y-axis = 0.

∴ P(0, -7) lies on y-axis.

Hence, Option 3 is the correct option.

Abscissa (or x-coordinate) is positive in I and IV quadrants.

Hence, Option 2 is the correct option.

Let point be P(x, y).

Since,

Point lies on y-axis.

∴ x = 0.

Also, point is at a distance of 5 units in the negative direction of y-axis.

∴ y = -5.

P = (x, y) = (0, -5).

Hence, Option 3 is the correct option.

In graph, let 1 block = 1 unit.

Since,

Perpendicular distance of a point P from the x-axis is 5 units and the foot of perpendicular lies on the negative direction of x-axis.

Let foot of perpendicular be B.

So, position of point P can be A or C.

From graph,

The y-coordinate of A and C = 5 and -5.

Hence, Option 4 is the correct option.

In second quadrant,

x-coordinate is negative and y-coordinate is positive.

In fourth quadrant,

x-coordinate is positive and y-coordinate is negative.

Hence, points having opposite signs lie in II and IV quadrants.

Hence, Option 4 is the correct option.

In second quadrant,

x-coordinate is negative and y-coordinate is positive.

∴ (-5, 2) lies in II quadrant.

In fourth quadrant,

x-coordinate is positive and y-coordinate is negative.

∴ (2, -5) lies in IV quadrant.

Hence, Option 3 is the correct option.

From graph,

Q and R lies in fourth quadrant.

Hence, Option 2 is the correct option.

From graph,

OA = BC = 3 units

OC = AB = 4 units

∴ Figure obtained is a rectangle.

Hence, Option 2 is the correct option.

Substituting (1, -2) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(1) - 5(-2) + 7

⇒ 3 + 10 + 7

⇒ 20

Since, L.H.S. ≠ R.H.S., (1, -2) does not satisfy the equation.

Substituting (2, 1) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(2) - 5(1) + 7

⇒ 6 - 5 + 7

⇒ 8

Since, L.H.S. ≠ R.H.S., (2, 1) does not satisfy the equation.

Substituting (-1, 2) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(-1) - 5(2) + 7

⇒ -3 - 10 + 7

⇒ -6

Since, L.H.S. ≠ R.H.S., (-1, 2) does not satisfy the equation.

Substituting (1, 2) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(1) - 5(2) + 7

⇒ 3 - 10 + 7

⇒ 0

Since, L.H.S. = R.H.S., (1, 2) satisfies the equation.

Hence, Option 4 is the correct option.

From graph,

x = a and y = b intersects at (a, b).

Hence, Option 4 is the correct option.

The distance between the point P(2, 3) and x-axis can be determined by assuming a point A(2, 0) on x-axis.

By distance formula,

$$\begin{aligned}\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow AP = \sqrt{(2 - 2)^2 + (3 - 0)^2} \\[1em] \Rightarrow AP = \sqrt{0 + 3^2} \\[1em] \Rightarrow AP = \sqrt{9} \\[1em] \Rightarrow AP = 3 \text{ units}.\end{aligned}$$

Hence, Option 2 is the correct option.

The distance between the point P(-4, 3) and y-axis can be determined by assuming a point A(0, 3) on y-axis.

By distance formula,

$$\begin{aligned}\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow AP = \sqrt{(-4 - 0)^2 + (3 - 3)^2} \\[1em] \Rightarrow AP = \sqrt{(-4)^2 + 0} \\[1em] \Rightarrow AP = \sqrt{16} \\[1em] \Rightarrow AP = 4 \text{ units}.\end{aligned}$$

Hence, Option 3 is the correct option.

Origin (O) = (0, 0) and P = (-6, 8).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore OP = \sqrt{(-6 - 0)^2 + (8 - 0)^2} \\[1em] = \sqrt{(-6)^2 + 8^2} \\[1em] = \sqrt{36 + 64} \\[1em] = \sqrt{100} \\[1em] = 10 \text{ units}.\end{aligned}$$

Hence, Option 3 is the correct option.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(0 - 0)^2 + (-2 - 6)^2} \\[1em] = \sqrt{0 + (-8)^2} \\[1em] = \sqrt{64} \\[1em] = 8 \text{ units}.\end{aligned}$$

Hence, Option 2 is the correct option.

Let, A = (0, 5) and B = (-5, 0).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(-5 - 0)^2 + (0 - 5)^2} \\[1em] = \sqrt{(-5)^2 + (-5)^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}\text{ units}.\end{aligned}$$

Hence, Option 2 is the correct option.

Since, AOBC is the rectangle.

So, AB will be the diagonal.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(5 - 0)^2 + (0 - 3)^2} \\[1em] = \sqrt{(5)^2 + (-3)^2} \\[1em] = \sqrt{25 + 9} \\[1em] = \sqrt{34} \text{ units}.\end{aligned}$$

Hence, Option 3 is the correct option.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

Distance between (2, -2) and (-1, x) is 5 units.

$$\begin{aligned}\therefore 5 = \sqrt{(-1 - 2)^2 + [x - (-2)]^2} \\[1em] \Rightarrow 5 = \sqrt{(-3)^2 + [x + 2]^2} \\[1em] \Rightarrow 5 = \sqrt{9 + x^2 + 4 + 4x} \\[1em] \Rightarrow 5 = \sqrt{x^2 + 4x + 13} \\[1em] \Rightarrow x^2 + 4x + 13 = 5^2 \text{ [On squaring both sides]} \\[1em] \Rightarrow x^2 + 4x + 13 = 25 \\[1em] \Rightarrow x^2 + 4x - 12 = 0 \\[1em] \Rightarrow x^2 + 6x - 2x - 12 = 0 \\[1em] \Rightarrow x(x + 6) - 2(x + 6) = 0 \\[1em] \Rightarrow (x - 2)(x + 6) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -6.\end{aligned}$$

As distance cannot be negative,

∴ x = 2

Hence, Option 2 is the correct option.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

Distance between (4, p) and (1, 0) is 5 units.

$$\begin{aligned}\therefore 5 = \sqrt{(1 - 4)^2 + (0 - p)^2} \\[1em] \Rightarrow 5 = \sqrt{(-3)^2 + (-p)^2} \\[1em] \Rightarrow 5 = \sqrt{9 + p^2} \\[1em] \Rightarrow p^2 + 9 = 5^2 \text{ [On squaring both sides]} \\[1em] \Rightarrow p^2 + 9 = 25 \\[1em] \Rightarrow p^2 = 16 \\[1em] \Rightarrow p = \sqrt{16} \\[1em] \Rightarrow p = \pm 4.\end{aligned}$$

Hence, Option 3 is the correct option.

Let A = (-4, 0), B = (4, 0) and C = (0, 3).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{[4 - (-4)]^2 + (0 - 0)^2} \\[1em] = \sqrt{[4 + 4]^2 + 0^2} \\[1em] = \sqrt{8^2} \\[1em] = \sqrt{64} \\[1em] = 8 \text{ units} \\[1em] \therefore BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[1em] \therefore AC = \sqrt{[0 - (-4)]^2 + (3 - 0)^2} \\[1em] = \sqrt{4^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[1em]\end{aligned}$$

Since, AC = BC.

∴ ABC is an isosceles triangle.

Hence, Option 2 is the correct option.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(3 - 0)^2 + [1 - (-2)]^2} \\[1em] = \sqrt{3^2 + [1 + 2]^2} \\[1em] = \sqrt{9 + 3^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}.\end{aligned}$$

Area of square = (side)² = AB²

= $(3\sqrt{2})^2$ = 18 sq. units.

Hence, Option 1 is the correct option.

Draw a perpendicular AD from A on x-axis.

From graph,

D = (1, 0).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AD = \sqrt{(1 - 1)^2 + (0 - 3)^2} \\[1em] = \sqrt{0 + (-3)^2} \\[1em] = \sqrt{9} \\[1em] = 3 \text{ units}. \\[1em] \therefore BC = \sqrt{[4 - (-1)]^2 + (0 - 0)^2} \\[1em] = \sqrt{[4 + 1]^2 + 0} \\[1em] = \sqrt{5^2} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.\end{aligned}$$

Area of triangle = $\frac{1}{2} \times$ base × height

= $\frac{1}{2} \times 3 \times 5$

= $\frac{15}{2}$ = 7.5 sq. units

Hence, Option 3 is the correct option.

Let A(0, 4), B(0, 0) and C(3, 0) be the vertices of the triangle.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(0 - 0)^2 + (0 - 4)^2} \\[1em] = \sqrt{0 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] \therefore BC = \sqrt{(3 - 0)^2 + (0 - 0)^2} \\[1em] = \sqrt{3^2 + 0} \\[1em] = \sqrt{9} \\[1em] = 3 \text{ units} \\[1em] \therefore AC = \sqrt{(3 - 0)^2 + (0 - 4)^2} \\[1em] = \sqrt{3^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = 5 \text{ units}.\end{aligned}$$

Perimeter = AB + BC + AC = 4 + 3 + 5 = 12 units.

Hence, Option 2 is the correct option.

Since, A is a point on y-axis, so x-coordinate = 0.

A = (0, 5).

B = (-3, 1).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(-3 - 0)^2 + (1 - 5)^2} \\[1em] = \sqrt{(-3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.\end{aligned}$$

Hence, Option 2 is the correct option.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(9 - 9)^2 + (6 - 0)^2} \\[1em] = \sqrt{0 + 6^2} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] \therefore BC = \sqrt{(-9 - 9)^2 + (6 - 6)^2} \\[1em] = \sqrt{(-18)^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units}. \\[1em] \therefore CD = \sqrt{[-9 - (-9)]^2 + (0 - 6)^2} \\[1em] = \sqrt{[-9 + 9]^2 + (-6)^2} \\[1em] = \sqrt{0 + 36} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] \therefore AD = \sqrt{(-9 - 9)^2 + (0 - 0)^2} \\[1em] = \sqrt{(-18)^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units}.\end{aligned}$$

Since, AB = CD and BC = AD.

∴ ABCD is a rectangle.

Hence, Option 1 is the correct option.

Any point that lies on the y-axis has its x-coordinate equal to zero.

The given point is (x², y).

For this point to lie on the y-axis, its x-coordinate, which is x², must be equal to zero.

So, x² = 0.

If x² = 0, then x must be 0.

∴ Statement 1 is true.

The abscissa is the x-coordinate of a point in a Cartesian coordinate system.

All points on the y-axis are of the form (0, y), where y can be any real number.

For any point on the y-axis, its x-coordinate (abscissa) is 0.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.

Two ordered pairs (a, b) and (c, d) are equal if a = c and b = d.

This statement is the definition of equality for ordered pairs. For two ordered pairs to be considered equal, their corresponding elements must be equal.

So, (a, b) = (c, d) if and only if a = c and b = d.

∴ Assertion (A) is true.

If a ≠ b, then (a, b) ≠ (b, a).

For example, the ordered pair (2, 3) is not the same as (3,2). They represent different points in a coordinate plane.

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

Point P(3, -5):

The x-coordinate is positive, and the y-coordinate is negative.

A point with a positive x-coordinate and a negative y-coordinate lies in quadrant IV.

Point Q(-2, -1):

The x-coordinate is negative, and the y-coordinate is negative.

A point with a negative x-coordinate and a negative y-coordinate lies in quadrant III.

∴ Assertion (A) is false.

Quadrant I: x-coordinate is positive, y-coordinate is positive (+, +).

Quadrant II: x-coordinate is negative, y-coordinate is positive (-, +).

Quadrant III: x-coordinate is negative, y-coordinate is negative (-, -).

Quadrant IV: x-coordinate is positive, y-coordinate is negative (+, -).

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

When a point (x,y) is reflected across the origin (0, 0), its new coordinates become (−x, −y). This transformation is also known as point symmetry with respect to the origin.

Point A has coordinates (3, 3).

If we reflect point A(3, 3) across the origin (0, 0), the coordinates of its reflected image would be (−3, −3).

The coordinates of the given point B are exactly (−3, −3).

Since point B is the exact reflection of point A across the origin, it means that the origin (0, 0) lies precisely on the straight line segment connecting point A and point B.

Therefore, the line segment AB passes through the origin.

∴ Assertion (A) is true.

Origin is the point of intersection of the coordinate axes.

This is the definition of the origin in a Cartesian coordinate system. It's the point where the x-axis and y-axis intersect, and its coordinates are (0, 0).

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

Given, the points A(2, 9), B(2, 5) and C(5, 5).

By distance formula,

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given, points A(2, 9), B(2, 5) and C(5, 5).

$$\begin{aligned}\text{Distance between AB }= \sqrt{(2 - 2)^2 + (5 - 9)^2} \\[1em] = \sqrt{0 + (-4)^2} \\[1em] = \sqrt{0 + 16} \\[1em] = \sqrt{16} \\[1em] = \text{4 units}. \\[1em] \text{Distance between BC } = \sqrt{(5 - 2)^2 + (5 - 5)^2} \\[1em] = \sqrt{3^2 + 0^2} \\[1em] = \sqrt{9 + 0} \\[1em] = \sqrt{9} \\[1em] = \text{3 units}. \\[1em] \text{Distance between AC } = \sqrt{(5 - 2)^2 + (5 - 9)^2} \\[1em] = \sqrt{3^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = \text{5 units}.\end{aligned}$$

AC² = 5² = 25 units

AB² = 4² = 16 units

BC² = 3² = 9 units

Since, AC² = AB² + BC², thus it satisfies pythagoras theorem.

Thus, ABC is a right angled triangle at B.

Thus, Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

A point lies on the y-axis if and only if its x-coordinate (abscissa) is 0.

∴ Reason (R) is true.

Let P = (0, 9), A = (6, 5), and B = (-4, 3).

By formula,

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}PA = \sqrt{(6 - 0)^2 + (5 - 9)^2}\\[1em] = \sqrt{6^2 + (-4)^2}\\[1em] = \sqrt{36 + 16}\\[1em] = \sqrt{52}. \\[1em] PB = \sqrt{(-4 - 0)^2 + (3 - 9)^2}\\[1em] = \sqrt{(-4)^2 + (-6)^2}\\[1em] = \sqrt{16 + 36}\\[1em] = \sqrt{52}\end{aligned}$$

Since PA = PB = $\sqrt{52}$.

​Therefore, the point (0, 9) is a point on y-axis which is equidistant from (6, 5) and (-4, 3).

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

In graph,

1 block = 1 unit.

Steps of construction :

1. Plot the points A(2, -1), B(2, 7) and C(4, 7) on graph.

2. Join AB and BC.

3. Measure AB. Draw a line segment CD, from point C parallel to y-axis.

4. Measure BC. Draw a line segment AD, from point A parallel to x-axis.

5. Mark F and G mid-point of BC and CD respectively.

6. Join AC and BD diagonals of rectangle.

7. Mark E as the point of intersection of diagonals.

From graph,

Coordinates of D = (4, -1).

(i) From graph,

F is the mid-point of BC and F = (3, 7).

Hence, coordinates of mid-point of BC = (3, 7).

(ii) From graph,

G is the mid-point of CD and G = (4, 3).

Hence, coordinates of mid-point of CD = (4, 3).

(iii) From graph,

E is the point of intersection of diagonals.

Hence, point of intersection of the diagonals = (3, 3).

From graph,

AB = 8 units and BC = 2 units.

Area of the rectangle ABCD = length × breadth

= AB × BC

= 8 × 2

= 16 sq. units.

Hence, the area of the rectangle = 16 sq. units.

Steps of construction :

1. Plot the points A(3, 5), B(3, -1) and C(-1, -3) on graph paper.

2. Join AB and BC.

3. From C draw a line CD parallel to AB, such that CD = AB.

4. From A draw a line AD parallel to BC, such that AD = BC.

5. Mark E, the mid-point of CD.

From graph,

The coordinates of fourth vertex D are (-1, 3).

The coordinates of the midpoint of CD i.e. E are (-1, 0).

As, 1 block = 1 unit

EF = 4 units

CD = 6 units

Area of parallelogram ABCD = Base × height

= CD × EF

= 6 × 4

= 24 sq. units.

Hence, D = (-1, 3), coordinates of mid-point of CD = (-1, 0) and the area of the parallelogram is 24 sq. units.

Given,

y = 2x - 1 ..........(1)

When x = 1, y = 2 × 1 - 1 = 2 - 1 = 1

x = 2, y = 2 × 2 - 1 = 4 - 1 = 3

x = 3, y = 2 × 3 - 1 = 6 - 1 = 5.

x	1	2	3
y	1	3	5

Steps of construction :

1. Plot the points (1, 1), (2, 3) and (3, 5) on graph.

2. Join the points.

Comparing equation (1) with y = mx + c, we get :

m = 2 and c = -1.

Hence, slope of the line = 2 and y-intercept = -1.

Given,

⇒ 2x + 3y = 6

⇒ 3y = 6 - 2x

⇒ y = $\frac{6}{3} - \frac{2x}{3}$

⇒ y = $-\frac{2x}{3} + 2$ ...........(1)

When x = 0, $y = 2 - \frac{2x}{3} = 2 - \frac{2 \times 0}{3}$ = 2 - 0 = 2,

x = 3, y = $2 - \frac{2 \times 3}{3}$ = 2 - 2 = 0,

x = 6, y = $2 - \frac{2 \times 6}{3} = 2 - \frac{12}{3}$ = 2 - 4 = -2.

x	0	3	6
y	2	0	-2

Steps of construction :

1. Plot the points (0, 2), (3, 0) and (6, -2) on graph.

2. Join the points.

Comparing equation (1) with y = mx + c, we get :

m = $-\frac{2}{3}$ and c = 2.

Hence the slope of the line = $-\frac{2}{3}$ and y intercept = 2.

Given,

⇒ 2x - 3y = 4

⇒ 3y = 2x - 4

⇒ y = $\frac{2x - 4}{3}$

⇒ y = $\frac{2}{3}x - \frac{4}{3}$ ........(1)

When x = -1, y = $\frac{2}{3} \times -1 - \frac{4}{3} = -\frac{2}{3} - \frac{4}{3} = -\frac{6}{3}$ = -2,

x = 2, y = $\frac{2}{3} \times 2 - \frac{4}{3} = \frac{4}{3} - \frac{4}{3}$ = 0,

x = 5, y = $\frac{2}{3} \times 5 - \frac{4}{3} = \frac{10}{3} - \frac{4}{3} = \frac{6}{3}$ = 2.

x	-1	2	5
y	-2	0	2

Steps of construction :

1. Plot the points (-1, -2), (2, 0) and (5, 2) on graph.

2. Join the points.

Comparing equation (1) with y = mx + c.

m = $\frac{2}{3}$ and c = $-\frac{4}{3}$.

Hence, the slope of line = $\frac{2}{3}$ and y intercept = $-\frac{4}{3}$.

Given,

⇒ 3x - 4y = 12

⇒ 4y = 3x - 12

⇒ y = $\frac{3x - 12}{4}$

⇒ y = $\frac{3}{4}x - 3$.

When x = 0, y = $\frac{3}{4} \times 0 - 3$ = 0 - 3 = -3,

x = 4, y = $\frac{3}{4} \times 4 - 3$ = 3 - 3 = 0,

x = 8, y = $\frac{3}{4} \times 8 - 3 = \frac{24}{4} - 3$ = 6 - 3 = 3.

x	0	4	8
y	-3	0	3

Steps of construction :

1. Plot the points (0, -3), (4, 0) and (8, 3) on graph.

2. Join the points.

(i) Steps of construction :

1. Take a point N (x = -4) and draw a line parallel to y-axis, touching the graph at point M.

2. From M draw a line parallel to x-axis touching y-axis at point O (y = -6).

Hence, when x = -4, the value of y is -6.

(ii) Steps of construction :

1. Take a point P (y = 3) and draw a line parallel to x-axis, touching the graph at point R.

2. From R draw a line parallel to y-axis touching x-axis at point Q (x = 8).

Hence, when y = 3, the value of x is 8.

Given,

⇒ 2x - 3y = 7

⇒ 3y = 2x - 7

⇒ y = $\frac{2x - 7}{3}$ .........(1)

When, x = -1, y = $\frac{2 \times -1 - 7}{3} = \frac{-2 - 7}{3} = -\frac{9}{3}$ = -3,

x = 2, y = $\frac{2 \times 2 - 7}{3} = \frac{4 - 7}{3} = -\frac{3}{3}$ = -1,

x = 5, y = $\frac{2 \times 5 - 7}{3} = \frac{10 - 7}{3} = \frac{3}{3}$ = 1.

Table of values of equation (1) :

x	-1	2	5
y	-3	-1	1

Steps of construction :

1. Plot the points (-1, -3), (2, -1) and (5, 1) on graph.

2. Join the points.

Given,

⇒ x + 6y = 11

⇒ 6y = 11 - x

⇒ y = $\frac{11 - x}{6}$ ..........(2)

When x = -7, y = $\frac{11 - (-7)}{6} = \frac{18}{6}$ = 3,

x = -1, y = $\frac{11 - (-1)}{6} = \frac{12}{6}$ = 2,

x = 5, y = $\frac{11 - 5}{6} = \frac{6}{6}$ = 1.

Table of values of equation (2) :

x	-7	-1	5
y	3	2	1

Steps of construction :

1. Plot the points (-7, 3), (-1, 2) and (5, 1) on graph.

2. Join the points.

From graph,

The two lines intersect at P(5, 1).

Hence x = 5, y = 1.

Given,

⇒ x - 2y - 4 = 0

⇒ 2y = x - 4

⇒ y = $\frac{x - 4}{2}$ .........(1)

When x = -2, y = $\frac{-2 - 4}{2} = -\frac{6}{2}$ = -3,

x = 0, y = $\frac{0 - 4}{2} = -\frac{4}{2} = -2$,

x = 2, y = $\frac{2 - 4}{2} = -\frac{2}{2}$ = -1.

Table of values of equation (1) :

x	-2	0	2
y	-3	-2	-1

Steps of construction :

1. Plot the points (-2, -3), (0, -2) and (2, -1) on graph.

2. Join the points.

Given,

⇒ 2x + y - 3 = 0

⇒ y = 3 - 2x .........(2)

When x = 0, y = 3 - 2(0) = 3 - 0 = 3,

x = 1, y = 3 - 2(1) = 3 - 2 = 1,

x = 2, y = 3 - 2(2) = 3 - 4 = -1.

Table of values of equation (2) :

x	0	1	2
y	3	1	-1

Steps of construction :

1. Plot the points (0, 3), (1, 1) and (2, -1) on graph.

2. Join the points.

From graph,

The lines intersect each other at P(2, -1).

Hence, x = 2, y = -1.

(i) Given,

⇒ 6y = 5x + 10

⇒ y = $\frac{5x + 10}{6}$ .........(1)

When x = 1, y = $\frac{5 \times 1 + 10}{6} = \frac{15}{6}$ = 2.5

x = -2, y = $\frac{5 \times -2 + 10}{6} = \frac{-10 + 10}{6} = \frac{0}{6}$ = 0,

x = 4, y = $\frac{5 \times 4 + 10}{6} = \frac{20 + 10}{6} = \frac{30}{6}$ = 5.

Table of values of equation (1) :

x	1	-2	4
y	2.5	0	5

Steps of construction :

1. Plot the points (1, 2.5), (-2, 0) and (4, 5) on graph.

2. Join the points.

Given,

y = 5x - 15

When x = 2.5, y = 5 × 2.5 - 15 = 12.5 - 15 = -2.5,

x = 3, y = 5 × 3 - 15 = 15 - 15 = 0,

x = 4, y = 5 × 4 - 15 = 20 - 15 = 5.

Table of values of equation (2) :

x	2.5	3	4
y	-2.5	0	5

Steps of construction :

1. Plot the points (2.5, -2.5), (3, 0) and (4, 5) on graph.

2. Join the points.

From graph,

The lines intersect at point P(4, 5).

Hence, x = 4, y = 5.

(ii) From graph,

Triangle = PQR.

Draw a line PJ, from P perpendicular to x-axis.

PJ = 5 units

QR = 5 units

Area of triangle = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × QR × PJ

= $\frac{1}{2}$ × 5 × 5

= $\frac{25}{2}$

= 12.5 sq. units.

Hence, area = 12.5 sq. units.

Given,

⇒ 8y - 3x + 7 = 0

⇒ 8y = 3x - 7

⇒ y = $\frac{3x - 7}{8}$ .........(1)

When x = -3, y = $\frac{3 \times -3 - 7}{8} = \frac{-9 - 7}{8} = -\frac{16}{8}$ = -2,

x = 1, y = $\frac{3 \times 1 - 7}{8} = \frac{3 - 7}{8} = -\frac{4}{8}$ = -0.5

x = 5, y = $\frac{3 \times 5 - 7}{8} = \frac{15 - 7}{8} = \frac{8}{8}$ = 1.

Table of values of equation (1) :

x	-3	1	5
y	-2	-0.5	1

Steps of construction :

1. Plot the points (-3, -2), (1, -0.5), (5, 1) on graph.

2. Join the points.

Given,

⇒ 2x - y + 4 = 0

⇒ y = 2x + 4

When x = -2, y = 2(-2) + 4 = -4 + 4 = 0,

x = -1, y = 2(-1) + 4 = -2 + 4 = 2,

x = 0, y = 2(0) + 4 = 0 + 4 = 4.

Table of values of equation (2) :

x	-2	-1	0
y	0	2	4

Steps of construction :

1. Plot the points (-2, 0), (-1, 2), (0, 4) on graph.

2. Join the points.

Given,

⇒ 5x + 4y = 29

⇒ 4y = 29 - 5x

⇒ y = $\frac{29 - 5x}{4}$ ..........(3)

When x = 1, y = $\frac{29 - 5 \times 1}{4} = \frac{29 - 5}{4} = \frac{24}{4}$ = 6,

x = 3, y = $\frac{29 - 5 \times 3}{4} = \frac{29 - 15}{4} = \frac{14}{4}$ = 3.5,

x = 5, y = $\frac{29 - 5 \times 5}{4} = \frac{29 - 25}{4} = \frac{4}{4}$ = 1.

Table of values of equation (3) :

x	1	3	5
y	6	3.5	1

Steps of construction :

1. Plot the points (1, 6), (3, 3.5), (5, 1) on graph.

2. Join the points.

From graph,

These three lines intersect each other at (-3, -2), (5, 1) and (1, 6).

Hence, the coordinates of the vertices of the triangle formed by these lines are (-3, -2), (5, 1) and (1, 6).

Given,

⇒ y - 2 = 0

⇒ y = 2 ........(1)

Given,

⇒ 2y + x = 0

⇒ 2y = -x

⇒ y = -$\frac{x}{2}$ ............(2)

When, x = -2, y = $-\frac{-2}{2}$ = 1,

x = 0, y = $-\frac{0}{2}$ = 0,

x = 2, y = $-\frac{2}{2}$ = -1.

Table of equation (2) :

x	-2	0	2
y	1	0	-1

Steps of construction :

1. Plot the points (-2, 1), (0, 0), (2, -1) on graph.

2. Join the points.

Given,

⇒ y + 1 = 3(x - 2)

⇒ y + 1 = 3x - 6

⇒ y = 3x - 6 - 1

⇒ y = 3x - 7 ...........(3)

When x = 1, y = 3 × 1 - 7 = 3 - 7 = -4,

x = 2, y = 3 × 2 - 7 = 6 - 7 = -1,

x = 3, y = 3 × 3 - 7 = 9 - 7 = 2.

Table of equation (3) :

x	1	2	3
y	-4	-1	2

Steps of construction :

1. Plot the points (1, -4), (2, -1), (3, 2) on graph.

2. Join the points.

From graph,

A(-4, 2), B(3, 2), C(2, -1) are the vertices of the triangle.

From C, draw CD perpendicular to AB.

As, 1 block = 1 unit.

AB = 7 units and CD = 3 units

Area of triangle = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × AB × CD

= $\frac{1}{2}$ × 7 × 3

= $\frac{21}{2}$ = 10.5 sq. units

Hence, coordinates of the vertices of the triangle are (-4, 2), (3, 2), (2, -1) and area = 10.5 sq. units.

Given,

Ordinate of the point on the other end = 9.

Let abscissa = x.

Given,

Distance between the two ends (-2, 3) and (x, 9) = 10 units.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore \sqrt{[x - (-2)]^2 + (9 - 3)^2} = 10 \\[1em] \Rightarrow \sqrt{[x + 2]^2 + 6^2} = 10 \\[1em] \Rightarrow \sqrt{x^2 + 4 + 4x + 36} = 10 \\[1em] \Rightarrow x^2 + 4x + 40 = 100\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow x^2 + 4x + 40 - 100 = 0 \\[1em] \Rightarrow x^2 + 4x - 60 = 0\\[1em] \Rightarrow x^2 + 10x - 6x - 60 = 0 \\[1em] \Rightarrow x(x + 10) - 6(x + 10) = 0 \\[1em] \Rightarrow (x - 6)(x + 10) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x + 10 = 0 \\[1em] \Rightarrow x = 6 \text{ or } x = -10.\end{aligned}$$

Hence, abscissa of other end = 6 or -10.

It is given that

A(-4, -1), B(-1, 2) and C(α, 5) are the vertices of an isosceles triangle and AB is the unequal side.

∴ AC = BC.

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Since, AC = BC

$$\begin{aligned}\Rightarrow \sqrt{[α - (-4)]^2 + [5 - (-1)]^2} = \sqrt{[α - (-1)]^2 + [5 - 2]^2} \\[1em] \Rightarrow \sqrt{[α + 4]^2 + [5 + 1]^2} = \sqrt{[α + 1]^2 + [3]^2} \\[1em] \Rightarrow \sqrt{α^2 + 16 + 8α + 6^2} = \sqrt{α^2 + 1 + 2α + 9}\end{aligned}$$

On squaring both sides we get,

$$\begin{aligned}\Rightarrow α^2 + 16 + 8α + 6^2 = α^2 + 1 + 2α + 9 \\[1em] \Rightarrow α^2 + 8α + 16 + 36 = α^2 + 2α + 10 \\[1em] \Rightarrow α^2 - α^2 + 8α - 2α = 10 - 16 - 36 \\[1em] \Rightarrow 6α = -42 \\[1em] \Rightarrow α = -\frac{42}{6} \\[1em] \Rightarrow α = -7.\end{aligned}$$

Hence, α = -7.

Given,

A(-3, 2), B(α, β) and C(-1, 4) are the vertices of an isosceles triangle

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

As, AB = BC

$\therefore \sqrt{[α - (-3)]^2 + (β - 2)^2} = \sqrt{(-1 - α)^2 + (4 - β)^2}$

On squaring both sides,

$$\begin{aligned}\Rightarrow [α - (-3)]^2 + (β - 2)^2 = (-1 - α)^2 + (4 - β)^2 \\[1em] \Rightarrow [α + 3]^2 + (β - 2)^2 = (-1 - α)^2 + (4 - β)^2 \\[1em] \Rightarrow α^2 + 3^2 + 6α + β^2 + 4 - 4β = 1 + α^2 + 2α + 16 + β^2 - 8β \\[1em] \Rightarrow α^2 + 9 + 6α + β^2 + 4 - 4β = α^2 + 2α + 17 + β^2 - 8β \\[1em] \Rightarrow α^2 - α^2 + β^2 - β^2 + 6α - 2α - 4β + 8β = 17 - 9 - 4 \\[1em] \Rightarrow 4α + 4β = 4 \\[1em] \Rightarrow 4(α + β) = 4 \\[1em] \Rightarrow α + β = 1.\end{aligned}$$

Hence, proved that α + β = 1.

Let the points be A(3, 0), B(6, 4) and C(-1, 3).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(6 - 3)^2 + (4 - 0)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. \\[1em] \therefore BC = \sqrt{(-1 - 6)^2 + (3 - 4)^2} \\[1em] = \sqrt{(-7)^2 + (-1)^2} \\[1em] = \sqrt{49 + 1} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.\\[1em] \therefore AC = \sqrt{(-1 - 3)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5.\end{aligned}$$

∴ AB = AC = 5

∴ ΔABC is an isosceles triangle

AB² + AC² = 5² + 5²

= 25 + 25

= 50.

BC² = $(5\sqrt{2})^2$ = 50.

Since, AB² + AC² = BC².

Hence, proved that (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.

Let A(2, 1), B(0, 3), C(-2, 1) and D(0, -1) be the four points.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(0 - 2)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore BC = \sqrt{(-2 - 0)^2 + (1 - 3)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.\\[1em] \therefore CD = \sqrt{[0 - (-2)]^2 + [-1 - 1]^2} \\[1em] = \sqrt{[0 + 2]^2 + [-2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore AD = \sqrt{(0 - 2)^2 + (-1 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] =\sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.\end{aligned}$$

Since, AB = BC = CD = AD i.e. all sides are equal so, ABCD can be a rhombus or a square.

Calculating diagonals,

$$\begin{aligned}AC = \sqrt{(-2 - 2)^2 + (1 - 1)^2} \\[1em] =\sqrt{(-4)^2 + 0^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] BD = \sqrt{(0 - 0)^2 + (-1 - 3)^2} \\[1em] = \sqrt{0^2 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = \sqrt{4} \text{ units}.\end{aligned}$$

Since, diagonals are equal.

∴ ABCD is a square.

Area of square = (side)²

= (AB)²

= $(\sqrt{8})^2$

= 8 sq. units.

Hence, proved that (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square and area = 8 sq. units.

Let A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4).

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{[-5 - (-3)]^2 + [-5 - 2]^2} \\[1em] = \sqrt{[-5 + 3]^2 + [-7]^2} \\[1em] = \sqrt{[-2]^2 + [-7]^2} \\[1em] = \sqrt{4 + 49} \\[1em] = \sqrt{53}. \\[1em] \therefore BC = \sqrt{[2 - (-5)]^2 + [-3 - (-5)]^2} \\[1em] = \sqrt{[2 + 5]^2 + [-3 + 5]^2} \\[1em] = \sqrt{7^2 + 2^2} \\[1em] = \sqrt{49 + 4} \\[1em] = \sqrt{53}. \\[1em] \therefore CD = \sqrt{(4 - 2)^2 + [4 - (-3)]^2} \\[1em] = \sqrt{2^2 + 7^2} \\[1em] = \sqrt{4 + 49} \\[1em] = \sqrt{53}. \\[1em] \therefore AD = \sqrt{[4 - (-3)]^2 + (4 - 2)^2} \\[1em] = \sqrt{7^2 + 2^2} \\[1em] = \sqrt{49 + 4} \\[1em] = \sqrt{53}.\end{aligned}$$

Calculating diagonals,

$$\begin{aligned}AC = \sqrt{[2 - (-3)]^2 + [-3 - 2]^2} \\[1em] = \sqrt{[2 + 3]^2 + [-5]^2} \\[1em] = \sqrt{5^2 + [-5]^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2} \text{ units}. \\[1em] BD = \sqrt{[4 - (-5)]^2 + [4 - (-5)]^2} \\[1em] = \sqrt{[4 + 5]^2 + [4 + 5]^2} \\[1em] = \sqrt{9^2 + 9^2} \\[1em] = \sqrt{81 + 81} \\[1em] = \sqrt{162} \\[1em] = 9\sqrt{2} \text{ units}.\end{aligned}$$

Since, all sides are equal and diagonals are not equal.

∴ ABCD is a rhombus.

Area of rhombus = $\frac{1}{2} \times d_1 \times d_2$

$$\begin{aligned}= \frac{1}{2} \times 5\sqrt{2} \times 9\sqrt{2} \\[1em] = \frac{1}{2} \times 45 \times 2 \\[1em] = 45 \text{ sq. units}.\end{aligned}$$

Hence, ABCD is a rhombus and area = 45 sq. units.

Given,

Ends of a diagonal of a square are (-2, p) and (p, 2).

Area of square = 40 sq. units

By formula,

Area of square = (side)²

∴ (side)² = 40

⇒ side = $\sqrt{40} = 2\sqrt{10}$ units.

By formula,

Diagonal of a square = $\sqrt{2}$ × side = $\sqrt{2} \times 2\sqrt{10} = 2\sqrt{20}$.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{[p - (-2)]^2 + [2 - p]^2} \\[1em] \Rightarrow 2\sqrt{20} =\sqrt{[p + 2]^2 + [2 - p]^2} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{p^2 + 4 + 4p + 4 + p^2 - 4p} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{2p^2 + 8}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow 4 \times 20 = 2p^2 + 8 \\[1em] \Rightarrow 2p^2 = 80 - 8 \\[1em] \Rightarrow 2p^2 = 72 \\[1em] \Rightarrow p^2 = \frac{72}{2} \\[1em] \Rightarrow p^2 = 36 \\[1em] \Rightarrow p = \sqrt{36} \\[1em] \Rightarrow p = \pm 6.\end{aligned}$$

Hence, p = ±6.

By distance formula,

$$\begin{aligned}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(7 - 2)^2 + [3 - (-2)]^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50}. \\[1em] \therefore BC = \sqrt{(11 - 7)^2 + (-1 - 3)^2} \\[1em] = \sqrt{4^2 + (-4)^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32}. \\[1em] \therefore CD = \sqrt{(6 - 11)^2 + [-6 - (-1)]^2} \\[1em] = \sqrt{(-5)^2 + (-5)^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50}. \\[1em] \therefore AD = \sqrt{(6 - 2)^2 + [-6 - (-2)]^2} \\[1em] = \sqrt{4^2 + (-4)^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32}.\end{aligned}$$

Hence,

AB = CD and BC = AD

Since, opposite sides are equal,

Hence, proved that ABCD is a rectangle.

Let O(x, y) be the coordinates of the centre of the circle. Points A(5, 1), B(-3, -7), and C(7, -1) are on the circle.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

As, OA = OB [∵ Both are radius of the same circle]

$$\begin{aligned}\therefore \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{[x - (-3)]^2 + [y - (-7)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 25 - 10x + y^2 + 1 - 2y} = \sqrt{[x + 3]^2 + [y + 7]^2} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 10x - 2y + 26} = \sqrt{x^2 + 9 + 6x + y^2 + 49 + 14y} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 10x - 2y + 26} = \sqrt{x^2 + y^2 + 6x + 14y + 58}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow x^2 + y^2 - 10x - 2y + 26 = x^2 + y^2 + 6x + 14y + 58 \\[1em] \Rightarrow x^2 - x^2 + y^2 - y^2 + 6x + 10x + 14y + 2y = 26 - 58 \\[1em] \Rightarrow 16x + 16y = -32 \\[1em] \Rightarrow 16(x + y) = -32 \\[1em] \Rightarrow x + y = \frac{-32}{16} \\[1em] \Rightarrow x + y = -2 \\[1em] \Rightarrow x = -2 - y \space .........(1).\end{aligned}$$

As, OC = OB [∵ Both are radius of the same circle]

$$\begin{aligned}\therefore \sqrt{(x - 7)^2 + [y - (-1)]^2} = \sqrt{[x - (-3)]^2 + [y - (-7)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 49 - 14x + [y + 1]^2} = \sqrt{[x + 3]^2 + [y + 7]^2} \\[1em] \Rightarrow \sqrt{x^2 + 49 - 14x + y^2 + 1 + 2y} = \sqrt{x^2 + 9 + 6x + y^2 + 49 + 14y} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 14x + 2y + 50} = \sqrt{x^2 + y^2 + 6x + 14y + 58}\end{aligned}$$

On squaring both sides,

$$\begin{aligned}\Rightarrow x^2 + y^2 - 14x + 2y + 50 = x^2 + y^2 + 6x + 14y + 58 \\[1em] \Rightarrow x^2 + y^2 - x^2 - y^2 + 6x + 14x + 14y - 2y = 50 - 58 \\[1em] \Rightarrow 20x + 12y = -8 \\[1em] \Rightarrow 5x + 3y = -2 \\[1em]\end{aligned}$$

Substituting value of x in above equation from equation 1 :

⇒ 5(-2 - y) + 3y = -2

⇒ -10 - 5y + 3y = -2

⇒ -2y = -2 + 10

⇒ -2y = 8

⇒ y = -4.

⇒ x = -2 - y = -2 - (-4) = -2 + 4 = 2.

C = (x, y) = (2, -4).

Hence, the coordinates of the centre of the circle are (2, -4).