Distance Formula

Let (7, -5) = (x₁, y₁) and (3, -1) = (x₂, y₂)

Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(3 - 7)^2 + ((-1) - (-5))^2}\\[1em] = \sqrt{(-4)^2 + (4)^2}\\[1em] = \sqrt{16 + 16}\\[1em] = \sqrt{32}\\[1em] = 4{\sqrt2} \text{units}\end{aligned}$$

Hence, option 3 is the correct option.

Since, distance between origin and (x, y) = $\sqrt{x^2 + y^2}$

∴ Distance between origin and the point (-4, 3)

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

Hence, option 1 is the correct option.

Let (-3, 2) = (x₁, y₁) and (x, 10) = (x₂, y₂)

Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] 10 = \sqrt{(x - (-3))^2 + (10 - 2)^2}\\[1em] ⇒ 10 = \sqrt{(x + 3)^2 + 8^2}\\[1em] ⇒ 10^2 = (x + 3)^2 + (8)^2\\[1em] ⇒ 100 = (x + 3)^2 + 64\\[1em] ⇒ 100 - 64 = (x + 3)^2\\[1em] ⇒ 36 = (3 + x)^2\\[1em] ⇒ \sqrt{36} = 3 + x\\[1em] ± 6 = 3 + x\\[1em] 6 = x + 3 \text{ or} -6 = 3 + x \\[1em] x = 6 - 3 \text{ or} -6 -3 = x \\[1em] x = 3 \text{ or} -9\\[1em]\end{aligned}$$

Hence, option 3 is the correct option.

Given (x, y) is equidistant from (3, 6) and (-3, 4)

i.e. distance between (x, y) and (3, 6) = distance between (x, y) and (-3, 4)

$$\begin{aligned}\sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x - (-3))^2 + (y - 4)^2}\\[1em] ⇒ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2}\\[1em] ⇒ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2\\[1em] ⇒ x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 -8y\\[1em] ⇒ 36 - 16 = 6x + 6x + 12y -8y\\[1em] ⇒ 20 = 12x + 4y\\[1em] ⇒ 3x + y = 5\end{aligned}$$

Hence, option 4 is the correct option.

Let (0, 4) = (x₁, y₁), (0, 0) = (x₂, y₂) and (3, 0) = (x₃, y₃)

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

The perimeter of a triangle

$$\begin{aligned}= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} + \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} + \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}\\[1em] = \sqrt{(0 - 0)^2 + (0 - 4)^2} + \sqrt{(3 - 0)^2 + (0 - 0)^2} + \sqrt{(3 - 0)^2 + (0 - 4)^2}\\[1em] = \sqrt{(- 4)^2} + \sqrt{(3)^2} + \sqrt{(3)^2 + (- 4)^2}\\[1em] = \sqrt{16} + \sqrt{9} + \sqrt{9 + 16}\\[1em] = 4 + 3 + \sqrt{25}\\[1em] = 4 + 3 + 5\\[1em] = 12 \text{units}\end{aligned}$$

Hence, option 1 is the correct option.

(i) Let (-3, 6) = (x₁, y₁) and (2, -6) = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(2 - (-3))^2 + (-6 - 6)^2}\\[1em] = \sqrt{(5)^2 + (-12)^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = 13\end{aligned}$$

Hence, distance between the given points is 13.

(ii) Let (-a, -b) = (x₁, y₁) and (a, b) = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(a - (-a))^2 + (b - (-b))^2}\\[1em] = \sqrt{(2a)^2 + (2b)^2}\\[1em] = \sqrt{4a^2 + 4b^2}\\[1em] = 2\sqrt{a^2 + b^2}\\[1em]\end{aligned}$$

Hence, distance between the given points is $2\sqrt{a^2 + b^2}$.

(iii) Let $\Big(\frac{3}{5},2\Big)$ = (x₁, y₁) and $\Big(-\frac{1}{5}, 1\frac{2}{5}\Big)$ = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{\Big(\frac{-1}{5} - \frac{3}{5} \Big)^2 + \Big(\Big(1\frac{2}{5} - 2 \Big)\Big)^2}\\[1em] = \sqrt{\Big(\frac{-4}{5}\Big)^2 + \Big(\frac{7-10}{5}\Big)^2}\\[1em] = \sqrt{\frac{16}{25} + \Big(\frac{-3}{5}\Big)^2}\\[1em] = \sqrt{\frac{16}{25} + \frac{9}{25}}\\[1em] = \sqrt{\frac{25}{25}}\\[1em] = \sqrt{1}\\[1em] = 1\end{aligned}$$

Hence, distance between the given points is 1.

(iv) Let $\Big({\sqrt3 +1},1\Big)$ = (x₁, y₁) and (0, $\sqrt{3}$) = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(0 - ({\sqrt3 +1}))^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{(-{(\sqrt3 +1)})^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{3 + 1 + 2\sqrt3 + 1 + 3 - 2\sqrt3}\\[1em] = \sqrt{8}\\[1em] = 2\sqrt2\end{aligned}$$

Hence, distance between the given points is $2\sqrt2$ = 2.83.

(i) Since, distance between origin and (x, y) = $\sqrt{x^2 + y^2}$

∴ Distance between origin and the point (-8, 6)

$$\begin{aligned}= \sqrt{(-8)^2 + 6^2}\\[1em] = \sqrt{64 + 36}\\[1em] = \sqrt{100}\\[1em] = \text{10}\end{aligned}$$

Hence, the distance between the origin and the point (-8, 6) is 10.

(ii) Since, distance between origin and (x, y) = $\sqrt{x^2 + y^2}$

∴ Distance between origin and the point (-5, -12)

$$\begin{aligned}= \sqrt{(-5)^2 + (-12)^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = \text{13}\end{aligned}$$

Hence, the distance between the origin and the point (-5, -12) is 13.

(iii) Since, distance between origin and (x, y) = $\sqrt{x^2 + y^2}$

∴ Distance between origin and the point (8, -15)

$$\begin{aligned}= \sqrt{8^2 + (-15)^2}\\[1em] = \sqrt{64 + 225}\\[1em] = \sqrt{289}\\[1em] = \text{17}\end{aligned}$$

Hence, the distance between the origin and the point is 17.

Let (3, 1) = (x₁, y₁) and (0, x) = (x₂, y₂)

⇒ Distance between the given points =

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

Hence, the value of x = 5 and -3.

Let the co-ordinates of the point on the x-axis be (x, 0).

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

Let (x, 0) = (x₁, y₁) and (11, -8) = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8 - 0)^2}\\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8)^2}\\[1em] ⇒ 17^2 = (11 - x)^2 + (-8)^2\\[1em] ⇒ 289 = 121 + x^2 - 22x + 64\\[1em] ⇒ 289 = 185 + x^2 - 22x\\[1em] ⇒ 185 + x^2 - 22x - 289 = 0\\[1em] ⇒ x^2 - (26x - 4x) - 104 = 0\\[1em] ⇒ x^2 - 26x + 4x - 104 = 0\\[1em] ⇒ (x^2 - 26x) + (4x - 104) = 0\\[1em] ⇒ x(x - 26) + 4(x - 26) = 0\\[1em] ⇒ (x - 26)(x + 4) = 0\\[1em] ⇒ x = 26 \text{ and} -4\end{aligned}$$

Hence, the co-ordinates of points are (26, 0) and (-4, 0).

Let (0, y) = (x₁, y₁) and (-8, 4) = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 10 = \sqrt{(-8 - 0)^2 + (4 - y)^2}\\[1em] ⇒ 10 = \sqrt{(-8)^2 + (4 - y)^2}\\[1em] ⇒ 10^2 = 64 + 16 + y^2 - 8y\\[1em] ⇒ 100 = 80 + y^2 - 8y\\[1em] ⇒ 80 + y^2 - 8y - 100 = 0 \\[1em] ⇒ y^2 - 8y - 20 = 0 \\[1em] ⇒ y^2 - (10y - 2y) - 20 = 0 \\[1em] ⇒ y^2 - 10y + 2y - 20 = 0 \\[1em] ⇒ (y^2 - 10y) + (2y - 20) = 0 \\[1em] ⇒ y(y - 10) + 2(y - 10) = 0 \\[1em] ⇒ (y - 10)(y + 2) = 0 \\[1em] ⇒ y = 10 \text{ and } -2\end{aligned}$$

Hence, the co-ordinates of points are (0, 10) and (0, -2).

Let (a, 2a) = (x₁, y₁) and (4, 3) = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ \sqrt{10} = \sqrt{(4 - a)^2 + (3 - 2a)^2}\\[1em] ⇒ 10 = (4 - a)^2 + (3 - 2a)^2\\[1em] ⇒ 10 = 16 + a^2 - 8a + 9 + 4a^2 - 12a\\[1em] ⇒ 10 = 25 + 5a^2 - 20a\\[1em] ⇒ 25 + 5a^2 - 20a - 10 = 0\\[1em] ⇒ 5a^2 - 20a + 15 = 0\\[1em] ⇒ a^2 - 4a + 3 = 0\\[1em] ⇒ a^2 - (3a + 1a) + 3 = 0\\[1em] ⇒ a^2 - 3a - 1a + 3 = 0\\[1em] ⇒ a(a - 3) - 1(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 1) = 0\\[1em] ⇒ a = 3 \text{ and } 1\end{aligned}$$

For each value of a = x, we can find the corresponding value of y:

• If a = 3, then y = 2a = 6
• If a = 1, then y = 2a = 2

Hence, the co-ordinates of point A are (3, 6) and (1, 2).

Given point (2, -1) is equidistant from (a, 7) and (-3, a).

i.e. distance between (2, -1) and (a, 7) = distance between (2, -1) and (-3, a)

$$\begin{aligned}\sqrt{(a - 2)^2 + (7 - (-1))^2} = \sqrt{((-3) - 2)^2 + (a - (-1))^2}\\[1em] ⇒ \sqrt{(a - 2)^2 + (7 + 1)^2} = \sqrt{(-3 - 2)^2 + (a + 1)^2}\\[1em] ⇒ \sqrt{(a - 2)^2 + (8)^2} = \sqrt{(-5)^2 + (a + 1)^2}\\[1em] ⇒ (a - 2)^2 + (8)^2 = (-5)^2 + (a + 1)^2\\[1em] ⇒ a^2 + 4 - 4a + 64 = 25 + a^2 + 1 + 2a\\[1em] ⇒ a^2 - 4a + 68 = a^2 + 2a + 26\\[1em] ⇒ - 4a + 68 = 2a + 26\\[1em] ⇒ 68 - 26 = 2a + 4a\\[1em] ⇒ 6a = 42\\[1em] ⇒ a = \frac{42}{6}\\[1em] ⇒ a = 7\end{aligned}$$

Hence, the value of a = 7.

Let the required point on the x-axis be (x, 0).

Given (x, 0) is equidistant from (7, 6) and (-3, 4).

i.e. distance between (x, 0) and (7, 6) = distance between (x, 0) and (-3, 4)

$$\begin{aligned}\sqrt{(7 - x)^2 + (6 - 0)^2} = \sqrt{((-3) - x)^2 + (4 - 0)^2}\\[1em] ⇒ \sqrt{(7 - x)^2 + (6)^2} = \sqrt{(-3 - x)^2 + (4)^2}\\[1em] ⇒ (7 - x)^2 + (6)^2 = (-3 - x)^2 + (4)^2\\[1em] ⇒ 49 + x^2 - 14x + 36 = 9 + x^2 + 6x + 16\\[1em] ⇒ x^2 - 14x + 85 = x^2 + 6x + 25\\[1em] ⇒ - 14x + 85 = 6x + 25\\[1em] ⇒ - 14x - 6x = 25 - 85\\[1em] ⇒ - 20x = -60\\[1em] ⇒ x = \frac{60}{20}\\[1em] ⇒ x = 3\end{aligned}$$

Hence, the point on the x-axis which is equidistant from the points (7, 6) and (-3, 4) is (3, 0).

Let the required point on the y-axis be (0, y).

Given (0, y) is equidistant from (5, 2) and (-4, 3).

i.e. distance between (0, y) and (5, 2) = distance between (0, y) and (-4, 3)

$$\begin{aligned}\sqrt{(5 - 0)^2 + (2 - y)^2} = \sqrt{((-4) - 0)^2 + (3 - y)^2}\\[1em] ⇒ \sqrt{(5)^2 + (2 - y)^2} = \sqrt{(-4)^2 + (3 - y)^2}\\[1em] ⇒ (5)^2 + (2 - y)^2 = (-4)^2 + (3 - y)^2\\[1em] ⇒ 25 + 4 + y^2 - 4y = 16 + 9 + y^2 - 6y\\[1em] ⇒ 29 + y^2 - 4y = 25 + y^2 - 6y\\[1em] ⇒ 29 - 4y = 25 - 6y\\[1em] ⇒ - 4y + 6y = 25 - 29\\[1em] ⇒ 2y = - 4\\[1em] ⇒ y = - \frac{4}{2}\\[1em] ⇒ y = - 2\end{aligned}$$

Hence, the point on the y-axis which is equidistant from the points (5, 2) and (-4, 3) is (0, -2).

(i) The point P lies on the x-axis, so its co-ordinates are (x, 0).

The ordinate of point P is 0.

(ii) The point Q lies on the y-axis, so its co-ordinates are (0, y).

The abscissa of point Q is 0.

(iii) If the abscissa of point P is -12, then P = (-12, 0).

And if the ordinate of point Q is -16. So, then Q = (0, -16).

The length of the line segment PQ

$$\begin{aligned}= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(0 - (-12))^2 + ((-16) - 0)^2}\\[1em] = \sqrt{(-12)^2 + (-16)^2}\\[1em] = \sqrt{144 + 256}\\[1em] = \sqrt{400}\\[1em] = 20\end{aligned}$$

Hence, the length of PQ = 20.

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

The length of PQ

$$\begin{aligned}= \sqrt{(5 - 0)^2 + (10 - 5)^2}\\[1em] = \sqrt{5^2 + 5^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]\end{aligned}$$

The length of QR

$$\begin{aligned}= \sqrt{(6 - 5)^2 + (3 - 10)^2}\\[1em] = \sqrt{1^2 + (-7)^2}\\[1em] = \sqrt{1 + 49}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]\end{aligned}$$

The length of RP

$$\begin{aligned}= \sqrt{(6 - 0)^2 + (3 - 5)^2}\\[1em] = \sqrt{6^2 + (-2)^2}\\[1em] = \sqrt{36 + 4}\\[1em] = \sqrt{40}\\[1em] = 2\sqrt{10}\\[1em]\end{aligned}$$

PQ = QR ⇒ the triangle is isosceles triangle

Hence, the triangle PQR is an isosceles triangle.

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

The length of PQ

$$\begin{aligned}= \sqrt{(6 - 0)^2 + (2 - (-4))^2}\\[1em] = \sqrt{6^2 + 6^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em]\end{aligned}$$

The length of RS

$$\begin{aligned}= \sqrt{(-3 - 3)^2 + (-1 - 5)^2}\\[1em] = \sqrt{-6^2 + (-6)^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em]\end{aligned}$$

The length of QR

$$\begin{aligned}= \sqrt{(3 - 6)^2 + (5 - 2)^2}\\[1em] = \sqrt{(-3)^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em]\end{aligned}$$

The length of SP

$$\begin{aligned}= \sqrt{(-3 - 0)^2 + (-1 - (-4))^2}\\[1em] = \sqrt{-3^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em]\end{aligned}$$

PQ = RS
QR = SP

The length of diagonal QS =

$$\begin{aligned}= \sqrt{(-3 - 6)^2 + (-1 - 2)^2}\\[1em] = \sqrt{(-9)^2 + (-3)^2}\\[1em] = \sqrt{81 + 9}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em]\end{aligned}$$

The length of diagonal PR =

$$\begin{aligned}= \sqrt{(3 - 0)^2 + (5 - (-4))^2}\\[1em] = \sqrt{3^2 + 9^2}\\[1em] = \sqrt{9 + 81}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em]\end{aligned}$$

So, QS = PR

Since opposite sides are equal, and the diagonals are equal, we can conclude that the quadrilateral PQRS is a rectangle.

Hence, the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.

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

The length of AB

$$\begin{aligned}= \sqrt{(-3 - 1)^2 + (0 - (-3))^2}\\[1em] = \sqrt{(-4)^2 + 3^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5\end{aligned}$$

The length of BC

$$\begin{aligned}= \sqrt{(4 - (-3))^2 + (1 - 0)^2}\\[1em] = \sqrt{(-7)^2 + 1^2}\\[1em] = \sqrt{49 + 1}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]\end{aligned}$$

The length of CA

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

If ABC is an right angled triangle,

AB² + CA² = $(2\sqrt5)$² = 5² + 5² = 25 + 25 = 50 ⇒ BC²

BC² = AB² + CA² ⇒ the triangle is right angled triangle.

and,

AB = CA ⇒ the triangle is isosceles triangle.

Base of triangle = Height of the triangle = 5 units.

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

$$\begin{aligned}= \frac{1}{2} \times 5 \times 5\\[1em] = \frac{25}{2}\\[1em] = 12.5 \text{ sq. units}\end{aligned}$$

Hence, the triangle ABC is an isosceles right-angled triangle and area of the triangle = 12.5 sq. units.

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

The length of AB

$$\begin{aligned}= \sqrt{(1 - 5)^2 + (5 - 6)^2}\\[1em] = \sqrt{(-4)^2 + (-1)^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17}\end{aligned}$$

The length of BC

$$\begin{aligned}= \sqrt{(2 - 1)^2 + (1 - 5)^2}\\[1em] = \sqrt{1^2 + (-4)^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17}\end{aligned}$$

The length of CD

$$\begin{aligned}= \sqrt{(6 - 2)^2 + (2 - 1)^2}\\[1em] = \sqrt{4^2 + 1^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17}\end{aligned}$$

The length of DA

$$\begin{aligned}= \sqrt{(6 - 5)^2 + (2 - 6)^2}\\[1em] = \sqrt{1^2 + 4^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17}\end{aligned}$$

AB = BC = CD = DA = $\sqrt{17}$

The length of diagonal AC =

$$\begin{aligned}= \sqrt{(2 - 5)^2 + (1 - 6)^2}\\[1em] = \sqrt{(-3)^2 + (-5)^2}\\[1em] = \sqrt{9 + 25}\\[1em] = \sqrt{34}\end{aligned}$$

The length of diagonal BD =

$$\begin{aligned}= \sqrt{(6 - 1)^2 + (2 - 5)^2}\\[1em] = \sqrt{(-5)^2 + (-3)^2}\\[1em] = \sqrt{25 + 9}\\[1em] = \sqrt{34}\end{aligned}$$

So, AC = BD

Since all sides and diagonals are equal, the points form a square.

Hence, the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

Plot the point P(-8, 6) on the graph paper.

Draw a perpendicular line from the point P to the x-axis.

From graph, it is clear that the distance between the point (-8, 6) and the x-axis is 6 units.

Hence, option 2 is the correct option.

Origin (O) = (0, 0) and P = (-4, -3).

By distance formula,

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

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

Hence, option 4 is the correct option.

Since, OPQ is right angled triangle. Using pythagoras theorem,

⇒ Hypotenuse² = Base² + Height²

⇒ OP² = OQ² + QP²

⇒ 5² = 3² + QP²

⇒ 25 = 9 + QP²

⇒ QP² = 25 - 9

⇒ QP² = 16

⇒ QP = $\sqrt{16}$

⇒ QP = 4 units

Since, OQ = 3 units and QP = 4 units.

The co-ordinates of point P are (3, 4).

Hence, option 4 is the correct option.

Given, AB is diameter of a circle with center at point P = (x, 0) and point B = (0, y).

By distance formula,

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

From figure,

AB (Diameter) = 2 x PB (Radius)

$$\begin{aligned}\Rightarrow 10 = 2 \times \sqrt{(x - 0)^2 + (0 - y)^2}\\[1em] \Rightarrow \frac{10}{2} = \sqrt{x^2 + (-y)^2}\\[1em] \Rightarrow 5 = \sqrt{x^2 + y^2}\\[1em] \Rightarrow 5^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 25.\end{aligned}$$

Hence, option 4 is the correct option.

Given, point P, x = -4 and y = 3.

Using distance formula,

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

Distance of P(-4, 3) from origin (0, 0)

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

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Given, the point P(x, y) is at a distance of 6 unit from origin.

By distance formula,

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

$$\begin{aligned}\Rightarrow 6 = \sqrt{(x - 0)^2 + (y - 0)^2}\\[1em] \Rightarrow 6 = \sqrt{(x)^2 + (y)^2}\\[1em] \Rightarrow 6^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 36.\end{aligned}$$

The above equation defines a circle centered at the origin with radius 6. That circle covers all four quadrants, so P can lie anywhere on that circle, not necessarily in the first quadrant.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Given, A = (2x, y), B = (x, 2y).

By distance formula,

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

Since, AB = 5 unit.

Substituting the values, we get :

$$\begin{aligned}\Rightarrow 5 = \sqrt{(x - 2x)^2 + (2y - y)^2}\\[1em] \Rightarrow 5 = \sqrt{(-x)^2 + y^2}\\[1em] \Rightarrow 5^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 25.\end{aligned}$$

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Given, the distance between the points A(x, 2x) and B(x, 0) is 4 unit.

By distance formula,

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

Substituting the values, we get :

$$\begin{aligned}\Rightarrow 4 = \sqrt{(x - x)^2 + (2x - 0)^2}\\[1em] \Rightarrow 4 = \sqrt{0^2 + (2x)^2}\\[1em] \Rightarrow 4 = \sqrt{4x^2}\\[1em] \Rightarrow 4^2 = 4x^2\\[1em] \Rightarrow 16 = 4x^2\\[1em] \Rightarrow x^2 = \frac{16}{4}\\[1em] \Rightarrow x^2 = 4\\[1em] \Rightarrow x = \sqrt{4}\\[1em] \Rightarrow x = \pm 2.\end{aligned}$$

The coordinates of B = (x, 0) = (2, 0) or (-2, 0)

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Let the point on y-axis be (0, y).

Let (0, y) = (x₁, y₁) and (-4, 7) = (x₂, y₂)

⇒ Distance between the given points =

$$\begin{aligned}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4 - 0)^2 + (7 - y)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4)^2 + (7 - y)^2}\\[1em] ⇒ (2 \sqrt5)^2 = (-4)^2 + (7 - y)^2\\[1em] ⇒ 20 = 16 + 49 + y^2 - 14y\\[1em] ⇒ 20 = 65 + y^2 - 14y\\[1em] ⇒ 65 + y^2 - 14y - 20 = 0\\[1em] ⇒ y^2 - 14y + 45 = 0\\[1em] ⇒ y^2 - (9y + 5y) + 45 = 0\\[1em] ⇒ y^2 - 9y - 5y + 45 = 0\\[1em] ⇒ (y^2 - 9y) - (5y - 45) = 0\\[1em] ⇒ y(y - 9) - 5(y - 9) = 0\\[1em] ⇒ (y - 9)(y - 5) = 0\\[1em] ⇒ y = 9 \text{ or } 5\end{aligned}$$

Hence, the the points on the y-axis are (0, 9) or (0, 5).

Given (2, 4) is equidistant from (5, k) and (k, 7).

i.e. distance between (2, 4) and (5, k) = distance between (2, 4) and (k, 7)

$$\begin{aligned}\sqrt{(2 - 5)^2 + (4 - k)^2} = \sqrt{(2 - k)^2 + (4 - 7)^2}\\[1em] ⇒ (2 - 5)^2 + (4 - k)^2 = (2 - k)^2 + (4 - 7)^2\\[1em] ⇒ (- 3)^2 + (4 - k)^2 = (2 - k)^2 + (- 3)^2\\[1em] ⇒ 9 + 16 + k^2 - 8k = 4 + k^2 - 4k + 9\\[1em] ⇒ 25 + k^2 - 8k = 13 + k^2 - 4k\\[1em] ⇒ 13 + k^2 - 4k - 25 - k^2 + 8k = 0\\[1em] ⇒ - 12 + 4k = 0\\[1em] ⇒ 4k = 12\\[1em] ⇒ k = \frac{12}{4}\\[1em] ⇒ k = 3\end{aligned}$$

Hence, the value of k = 3.

Radius of circle = $\frac{\text{Diameter}}{2}$

= $\frac{10 \sqrt2}{2}$

= $5 \sqrt2$

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

Radius of the circle = The distance between the center (2a, a - 7) and the point (11, -9), which lies on the circle

$$\begin{aligned}5 \sqrt2 = \sqrt{(2a - 11)^2 + ((a - 7) - (-9))^2}\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a - 7 + 9)^2\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a + 2)^2\\[1em] ⇒ 50 = 4a^2 + 121 - 44a + a^2 + 4 + 4a\\[1em] ⇒ 50 = 5a^2 + 125 - 40a\\[1em] ⇒ 5a^2 + 125 - 40a - 50 = 0\\[1em] ⇒ 5a^2 - 40a + 75 = 0\\[1em] ⇒ a^2 - 8a + 15 = 0\\[1em] ⇒ a^2 - (3a + 5a) + 15 = 0\\[1em] ⇒ a^2 - 3a - 5a + 15 = 0\\[1em] ⇒ a(a - 3) - 5(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 5) = 0\\[1em] ⇒ a = 3 \text{ or } 5\end{aligned}$$

Hence, the value of a = 3 or a = 5.

The points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a quadrilateral. We need to show that this quadrilateral is a rhombus.

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

Distance between points A(-3, 2) and B(-5, -5):

$$\begin{aligned}= \sqrt{((-5) - (-3))^2 + ((-5) - 2)^2}\\[1em] = \sqrt{(-2)^2 + (-7)^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}\end{aligned}$$

Distance between points B(-5, -5) and C(2, -3):

$$\begin{aligned}= \sqrt{(2 - (-5))^2 + ((-3) - (-5))^2}\\[1em] = \sqrt{7^2 + 2^2}\\[1em] = \sqrt{49 + 4}\\[1em] = \sqrt{53}\end{aligned}$$

Distance between points C(2, -3) and D(4, 4):

$$\begin{aligned}= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}\end{aligned}$$

Distance between points D(4, 4) and A(-3, 2):

$$\begin{aligned}= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}\end{aligned}$$

AB = BC = CD = DA = $\sqrt{53}$

Since all sides are equal, the quadrilateral is a rhombus.

Hence, the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Given:

Points A(-3, -2), B(-6, a), C(-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD and AB = CD.

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

Distance between points A(-3, -2) and B(-6, a):

$$\begin{aligned}= \sqrt{((-6) - (-3))^2 + (a - (-2))^2}\\[1em] = \sqrt{(-3)^2 + (a - (-2))^2}\\[1em] = \sqrt{9 + a^2 + 4 + 4a}\\[1em] = \sqrt{13 + a^2 + 4a}\end{aligned}$$

Distance between points C (-3, -4) and D(0, -1):

$$\begin{aligned}= \sqrt{(0 - (-3))^2 + ((-1) - (-4))^2}\\[1em] = \sqrt{3^2 + 3^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\end{aligned}$$

Since AB = CD,

$$\begin{aligned}⇒ \sqrt{13 + a^2 + 4a} = \sqrt{18}\\[1em] ⇒ 13 + a^2 + 4a = 18\\[1em] ⇒ 13 + a^2 + 4a - 18 = 0\\[1em] ⇒ a^2 + 4a - 5 = 0\\[1em] ⇒ a^2 + 5a - 1a - 5 = 0\\[1em] ⇒ (a^2 + 5a) - (1a + 5) = 0\\[1em] ⇒ a(a + 5) - 1(a + 5) = 0\\[1em] ⇒ (a + 5)(a - 1) = 0\\[1em] ⇒ a = -5 \text{ or } 1\end{aligned}$$

Since it is given that a is negative, we select a = -5.

Hence, the value of a = -5.

Let the circumcentre of the triangle be P(x, y).

The circumcentre is equidistant from all three vertices of the triangle.

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

The distance from P to A is equal to the distance from P to B, so:

$$\begin{aligned}⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 1)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 1)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 1 - 2y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 122 - 22x + y^2 - 2y\\[1em] ⇒ 26 - 10x = 122 - 22x\\[1em] ⇒ 22x - 10x = 122 - 26\\[1em] ⇒ 12x = 96\\[1em] ⇒ x = \frac{96}{12}\\[1em] ⇒ x = 8\end{aligned}$$

The distance from P to A is equal to the distance from P to C, so:

$$\begin{aligned}⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 9)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 9)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 81 - 18y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 202 - 22x + y^2 - 18y\\[1em] ⇒ 26 - 10x - 2y = 202 - 22x - 18y\\[1em] ⇒ 26 - 10x - 2y - 202 + 22x + 18y = 0\\[1em] ⇒ 12x + 16y = 176\\[1em] ⇒ 3x + 4y = 44\\[1em]\end{aligned}$$

Putting the value of x = 8 in the above equation,

$$\begin{aligned}⇒ 3 \times 8 + 4y = 44\\[1em] ⇒ 24 + 4y = 44\\[1em] ⇒ 4y = 44 - 24\\[1em] ⇒ 4y = 20\\[1em] ⇒ y = \frac{20}{4}\\[1em] ⇒ y = 5\end{aligned}$$

Hence, the co-ordinates of the circumcentre of the triangle is (8, 5).

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

Distance between A(3, 1) and B(0, y - 1):

$$\begin{aligned}⇒ \sqrt{(0 - 3)^2 + ((y - 1) - 1)^2} = 5\\[1em] ⇒ (-3)^2 + (y - 1 - 1)^2 = 5^2\\[1em] ⇒ 9 + (y - 2)^2 = 25\\[1em] ⇒ 9 + y^2 + 4 - 4y = 25\\[1em] ⇒ y^2 - 4y + 13 = 25\\[1em] ⇒ y^2 - 4y - 25 + 13 = 0\\[1em] ⇒ y^2 - 4y - 12 = 0\\[1em] ⇒ y^2 - 6y + 2y - 12 = 0\\[1em] ⇒ (y^2 - 6y) + (2y - 12) = 0\\[1em] ⇒ y(y - 6) + 2(y - 6) = 0\\[1em] ⇒ (y - 6)(y + 2) = 0\\[1em] ⇒ y = 6 \text{ or } -2\end{aligned}$$

Hence, the values of y are 6 and -2.

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

Distance between A(x + 2, -2) and B(11, 6):

$$\begin{aligned}⇒ \sqrt{(11 - (x + 2))^2 + (6 - (-2))^2} = 17\\[1em] ⇒ (11 - x - 2)^2 + (6 + 2)^2 = 17^2\\[1em] ⇒ (9 - x)^2 + 8^2 = 289\\[1em] ⇒ 81 + x^2 - 18x + 64 = 289\\[1em] ⇒ x^2 - 18x + 145 = 289\\[1em] ⇒ x^2 - 18x + 145 - 289 = 0\\[1em] ⇒ x^2 - 18x - 144 = 0\\[1em] ⇒ x^2 - 24x + 6x - 144 = 0\\[1em] ⇒ (x^2 - 24x) + (6x - 144) = 0\\[1em] ⇒ x(x - 24) + 6(x - 24) = 0\\[1em] ⇒ (x - 24)(x + 6) = 0\\[1em] ⇒ x = 24 \text{ and } -6\end{aligned}$$

Hence, the values of x are 24 and -6.

The diameter of the circle is given as 20 units, so the radius is 10 units.

Distance between the centre A (2x - 1, 3x + 1) and point B (-3, -1) = Radius of circle AB = 10

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

⇒ AB² = 100

Distance between B(-3, -1) and A(2x - 1, 3x + 1):

$$\begin{aligned}⇒ ((2x - 1) - (-3))^2 + ((3x + 1) - (-1))^2 = 100\\[1em] ⇒ (2x - 1 + 3)^2 + (3x + 1 + 1)^2 = 100\\[1em] ⇒ (2x + 2)^2 + (3x + 2)^2 = 100\\[1em] ⇒ 4x^2 + 4 + 8x + 9x^2 + 4 + 12x = 100\\[1em] ⇒ 13x^2 + 8 + 20x = 100\\[1em] ⇒ 13x^2 + 8 + 20x - 100 = 0\\[1em] ⇒ 13x^2 + 20x - 92 = 0\\[1em] ⇒ x = \frac{-20 + \sqrt{400 + 4784}}{26} \text { or } \frac{-20 - \sqrt{400 + 4784}}{26}\\[1em] ⇒ x = \frac{-20 + \sqrt{5184}}{26} \text { or } \frac{-20 - \sqrt{5184}}{26}\\[1em] ⇒ x = \frac{-20 + 72}{26} \text { or } \frac{-20 - 72}{26}\\[1em] ⇒ x = \frac{52}{26} \text { or } \frac{-92}{26}\\[1em] ⇒ x = 2 \text { or } \frac{-46}{13}\\[1em]\end{aligned}$$

Hence, the values of x are 2 and $\frac{-46}{13}$.

Let the co-ordinates of point Q be (10, y).

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

Distance between P(2, -3) and Q(10, y):

$$\begin{aligned}⇒ \sqrt{(10 - 2)^2 + (y - (-3))^2} = 10\\[1em] ⇒ (10 - 2)^2 + (y - (-3))^2 = 100\\[1em] ⇒ 8^2 + (y + 3)^2 = 100\\[1em] ⇒ 64 + y^2 + 9 + 6y = 100\\[1em] ⇒ y^2 + 6y + 73 = 100\\[1em] ⇒ y^2 + 6y + 73 - 100 = 0\\[1em] ⇒ y^2 + 6y - 27 = 0\\[1em] ⇒ y^2 + 9y - 3y - 27 = 0\\[1em] ⇒ (y^2 + 9y) - (3y + 27) = 0\\[1em] ⇒ y(y + 9) - 3(y + 9) = 0\\[1em] ⇒ (y + 9)(y - 3) = 0\\[1em] ⇒ y = -9 \text{ or } 3\end{aligned}$$

Hence, the required co-ordinates of the point Q are (10, -9) and (10, 3).

(i) Given:

Radius = PA = PB = 13 units

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

Distance between P(2, -7) and T(-2, -4):

$$\begin{aligned}= \sqrt{(-2 - 2)^2 + (-4 + 7)^2}\\[1em] = \sqrt{(-4)^2 + (3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5\end{aligned}$$

Using Pythagoras theorem in triangle PAT,

PA² = PT² + AT²

⇒ AT² = PA² - PT²

⇒ AT² = 13² - 5²

⇒ AT² = 169 - 25

⇒ AT² = 144

⇒ AT = $\sqrt{144}$

⇒ AT = 12 units

Hence, the value of AT = 12 units.

(ii) We know that the perpendicular from the center of a circle to a chord bisects the chord.

AB = 2AT

= 2 x 12 units

= 24 units

Hence, the length of AB = 24 units.

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

Distance between P(2, 2) and Q(5, 4):

$$\begin{aligned}= \sqrt{(5 - 2)^2 + (4 - 2)^2}\\[1em] = \sqrt{3^2 + 2^2}\\[1em] = \sqrt{9 + 4}\\[1em] = \sqrt{13}\\[1em] = 3.605 \text{ units}\end{aligned}$$

Hence, the distance between the points P(2, 2) and Q(5, 4) is 3.605 units.

We know that any point on x-axis has co-ordinates of the form (x, 0).

The abscissa of point B is 11.

So, the point B is (11, 0).

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

Distance between A(7, 3) and B(11, 0):

$$\begin{aligned}= \sqrt{(11 - 7)^2 + (0 - 3)^2}\\[1em] = \sqrt{4^2 + (- 3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5 \text{ units}\end{aligned}$$

Hence, the distance between A(7, 3) and B(11, 0) is 5 units.

We know that any point on the y-axis has co-ordinates of the form (0, y).

The ordinate of point B is 9.

So, the point B is (0, 9).

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

Distance between A(5, -3) and B(0, 9):

$$\begin{aligned}= \sqrt{(0 - 5)^2 + (9 - (-3))^2}\\[1em] = \sqrt{(-5)^2 + 12^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = 13\end{aligned}$$

Hence, the distance between A(5, -3) and B(0, 9) is 13 units.

Let the point on y-axis be P(0, y).

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

Distance between A(6, 7) and P(0, y):

$$\begin{aligned}= \sqrt{(0 - 6)^2 + (y - 7)^2}\\[1em] = \sqrt{(- 6)^2 + (y - 7)^2}\\[1em] = \sqrt{36 + y^2 + 49 - 14y}\\[1em] = \sqrt{y^2 - 14y + 85}\\[1em]\end{aligned}$$

Distance between B(4, -3) and P(0, y):

$$\begin{aligned}= \sqrt{(0 - 4)^2 + (y - (-3))^2}\\[1em] = \sqrt{(- 4)^2 + (y + 3)^2}\\[1em] = \sqrt{16 + y^2 + 9 + 6y}\\[1em] = \sqrt{y^2 + 6y + 25}\\[1em]\end{aligned}$$

It is given that the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.

$$\begin{aligned}⇒\frac{PA}{PB} = \frac{1}{2}\\[1em] ⇒\frac{\sqrt{y^2 - 14y + 85}}{\sqrt{y^2 + 6y + 25}} = \frac{1}{2}\\[1em] ⇒\frac{y^2 - 14y + 85}{y^2 + 6y + 25} = \frac{1}{4}\\[1em] ⇒4(y^2 - 14y + 85) = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 - y^2 - 6y - 25 = 0\\[1em] ⇒ 3y^2 - 62y + 315 = 0\\[1em] ⇒ y = \frac{62 + \sqrt{3844 - 3780}}{6} \text { or } \frac{62 - \sqrt{3844 - 3780}}{6}\\[1em] ⇒ y = \frac{62 + \sqrt{64}}{6} \text { or } \frac{62 - \sqrt{64}}{6}\\[1em] ⇒ y = \frac{62 + 8}{6} \text { or } \frac{62 - 8}{6}\\[1em] ⇒ y = \frac{70}{6} \text { or } \frac{54}{6}\\[1em] ⇒ y = \frac{35}{3} \text { or } 9 \\[1em]\end{aligned}$$

Hence, the required points on y-axis are (0, 9) and $\Big(0, \frac{35}{3}\Big)$.

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

Distance between A(1, -3) and P(x, y):

$$\begin{aligned}= \sqrt{(x - 1)^2 + (y - (-3))^2}\\[1em] = \sqrt{(x - 1)^2 + (y + 3)^2}\\[1em] = \sqrt{x^2 + 1 - 2x + y^2 + 9 + 6y}\\[1em] = \sqrt{x^2 - 2x + y^2 + 10 + 6y}\\[1em]\end{aligned}$$

Distance between B(-2, 2) and P(x, y):

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

It is given that the distances of point P(x, y) from the points A(1, -3) and B(-2, 2) are in the ratio 2 : 3.

$$\begin{aligned}⇒\frac{PA}{PB} = \frac{2}{3}\\[1em] ⇒\frac{\sqrt{x^2 - 2x + y^2 + 10 + 6y}}{\sqrt{x^2 + 4x + y^2 + 8 - 4y}} = \frac{2}{3}\\[1em] ⇒\frac{x^2 - 2x + y^2 + 10 + 6y}{x^2 + 4x + y^2 + 8 - 4y} = \frac{4}{9}\\[1em] ⇒9(x^2 - 2x + y^2 + 10 + 6y) = 4(x^2 + 4x + y^2 + 8 - 4y)\\[1em] ⇒9x^2 - 18x + 9y^2 + 90 + 54y = 4x^2 + 16x + 4y^2 + 32 - 16y\\[1em] ⇒9x^2 - 18x + 9y^2 + 90 + 54y - 4x^2 - 16x - 4y^2 - 32 + 16y = 0\\[1em] ⇒ 5x^2 + 5y^2 - 34x + 70y + 58 = 0\\[1em]\end{aligned}$$

Hence, proved :- 5x² + 5y² - 34x + 70y + 58 = 0.

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

The length of AC:

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

The length of BC:

$$\begin{aligned}= \sqrt{(a - 4)^2 + ((-2) - (-1))^2}\\[1em] = \sqrt{(a - 4)^2 + (-1)^2}\\[1em] = \sqrt{a^2 + 16 - 8a + 1}\\[1em] = \sqrt{a^2 - 8a + 17}\end{aligned}$$

The length of AB:

$$\begin{aligned}= \sqrt{(a - 3)^2 + ((-2) - 0)^2}\\[1em] = \sqrt{(a - 3)^2 + (-2)^2}\\[1em] = \sqrt{a^2 + 9 - 6a + 4}\\[1em] = \sqrt{a^2 - 6a + 13}\end{aligned}$$

Using Pythagoras theorem in triangle ABC,

BC² = AB² + AC²

$$\begin{aligned}⇒ (\sqrt2)^2 + (\sqrt{a^2 - 6a + 13})^2 = (\sqrt{a^2 - 8a + 17})^2\\[1em] ⇒ 2 + a^2 - 6a + 13 = a^2 - 8a + 17\\[1em] ⇒ a^2 - 6a + 15 = a^2 - 8a + 17 \\[1em] ⇒ - 6a + 15 + 8a - 17 = 0\\[1em] ⇒ 2a - 2 = 0\\[1em] ⇒ 2a = 2 \\[1em] ⇒ a = \frac{2}{2} \\[1em] ⇒ a = 1\end{aligned}$$

Hence, the value of a = 1.

Given,

The co-ordinates are :

P(Nitin): (6, 4)

Q(Rajesh): (11, 9)

R(Kareem): (9, 7)

We know that

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

(i) (a) Distance between Nitin(P) and Rajesh(Q),

$$\begin{aligned}PQ = \sqrt{(11 - 6)^2 + (9 - 4)^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.\end{aligned}$$

Hence, distance between Nitin and Rajesh = $5\sqrt{2}$ units.

(b) Distance between Rajesh(Q) and Kareem(R),

$$\begin{aligned}QR = \sqrt{(11 - 9)^2 + (9 - 7)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.\end{aligned}$$

Hence, distance between Rajesh and Kareem = $2\sqrt{2}$ units.

(c) Distance between Nitin(P) and Kareem(R),

$$\begin{aligned}PR = \sqrt{(9 - 6)^2 + (7 - 4)^2} \\[1em] = \sqrt{3^2 + 3^2} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}.\end{aligned}$$

Hence, distance between Nitin and Kareem = $3\sqrt{2}$ units.

(ii) Points P, Q and R are collinear only if, PR + RQ = PQ

Consider L.H.S,

PR + RQ = $3\sqrt{2} + 2\sqrt{2}$

= $5\sqrt{2}$

Consider R.H.S,

PQ = $5\sqrt{2}$

∴ L.H.S = R.H.S

So, they are collinear.

Hence, P, Q and R are collinear.

(iii) Let S (x, 0) be the point on x-axis equidistant from Q and R.

According to question,

SQ = SR

$\sqrt{(x - 11)^2 + (0 - 9)^2} = \sqrt{(x - 9)^2 + (0 - 7)^2}$

Squaring on both sides,

(x - 11)² + (0 - 9)² = (x - 9)² + (0 - 7)²

x² - 22x + 121 + 81 = x² - 18x + 81 + 49

-18x + 22x = 121 + 81 - 81 - 49

4x = 121 - 49

4x = 72

x = $\frac{72}{4}$ = 18.

∴ The point S = (18, 0).

Hence, point on x-axis that is equidistant from Q and R = (18, 0).

(iv) If R is taken as origin, to find points of P and Q, their coordinates must be subtracted from the original coordinates of R(9, 7).

P' = (6 - 9, 4 - 7) = (-3, -3)

Q' = (11 - 9, 9 - 7) = (2, 2)

Hence, coordinates of P and Q are = (-3, -3) and (2, 2) respectively.

(i) Point A is on the X-axis at a distance of 10 km from O (0, 0). So, A = (10, 0).

Point B is on the Y-axis at a distance of 12 km from O (0, 0). So, B = (0, 12).

Hence, coordinates of A = (10, 0) and coordinates of B = (0, 12).

(ii) Using Pythagoras theorem for the △OAB:

AB² = BO² + OA²

AB² = 12² + 10²

AB² = 144 + 100

AB² = 244

AB = $\sqrt{244}$

AB = 15.6 km.

Hence, length of the local road AB = 15.6 km.

(iii) Based on the diagram, Sohan's village is at (30, 0) and Sita's village is at (0, 12).

$$\begin{aligned}\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(30 - 0)^2 + (0 - 12)^2} \\[1em] = \sqrt{30^2 + (-12)^2} \\[1em] = \sqrt{900 + 144} \\[1em] = \sqrt{1044} \\[1em] = 32.31 \text{ km}\end{aligned}$$

Hence, shortest distance between Sohan's village and Sita's village = 32.31 km.

(iv) Distance of Rohan's village from Gita's village:

Distance between Rohan's village to O + Distance between Gita's village to O

= 20 + 15

= 35 km.

Hence, distance from Rohan's village to Gita's village = 35 km.