The line segment joining A(2, -3) and B(-3, 2) is intercepted by the x-axis at the point M and the y-axis at the point N. PQ is perpendicular to AB at R and meets the y-axis at a distance of 6 units form the origin O, as shown in the diagram, at S. Find the :
(a) coordinates of M and N.
(b) coordinates of S
(c) slope of AB.
(d) equation of line PQ.
Show SolutionHide Solution↓
(a) From figure,
Coordinates of M = (-1, 0) and coordinates of N = (0, -1).
(b) From figure,
Coordinates of S = (0, 6).
(c) By formula,
Slope = $\frac{y_2 - y_1}{x_2 - x_1}$
Slope of AB = $\frac{2 - (-3)}{-3 - 2} = \frac{2 + 3}{-5} = \frac{5}{-5}$ = -1.
Hence, slope of AB = -1.
(d) We know that,
Product of slope of perpendicular lines = -1.
∴ Slope of AB × Slope of PQ = -1
⇒ -1 × Slope of PQ = -1
⇒ Slope of PQ = $\frac{-1}{-1}$ = 1.
Since, PQ passes through point S.
By point-slope form,
Equation of line : y - y1 = m(x - x1)
⇒ y - 6 = 1(x - 0)
⇒ y - 6 = x
⇒ y = x + 6.
Hence, equation of line PQ is y = x + 6.