A line segment joining P (2, -3) and Q (0, -1) is cut by the x-axis at the point R. A line AB cuts the y-axis at T(0, 6) and is perpendicular to PQ at S. Find the :

(a) equation of line PQ

(b) equation of line AB

(c) coordinates of points R and S.

(a) By formula,

Slope of line = $\Big(\dfrac{y$2 - y1}{x2 - x1}\Big)

Substituting values we get :

$\text{Slope of line PQ} = \Big(\dfrac{-1 - (-3)}{0 - 2}\Big) \\[1em] = \dfrac{-1 + 3}{-2} \\[1em] = \dfrac{2}{-2} \\[1em] = -1.$

Equation of line :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = -1(x - 2)

⇒ y + 3 = -x + 2

⇒ x + y + 3 - 2 = 0

⇒ x + y + 1 = 0.

Hence, equation of line PQ is x + y + 1 = 0.

(b) We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of PQ × Slope of AB = -1

⇒ -1 × Slope of AB = -1

⇒ Slope of AB = $\dfrac{-1}{-1}$ = 1.

Equation of line :

⇒ y - y₁ = m(x - x₁)

Equation of line AB :

⇒ y - 6 = 1(x - 0)

⇒ y - 6 = x

⇒ x - y + 6 = 0

Hence, equation of line AB is x - y + 6 = 0.

(c) Given,

Line PQ cuts x-axis at point R.

Let R = (a, 0)

Equation of line PQ = x + y + 1 = 0

Since, point R lies on line PQ,

⇒ a + 0 + 1 = 0

⇒ a + 1 = 0

⇒ a = -1.

R = (a, 0) = (-1, 0)

Given,

AB is perpendicular to PQ at point S.

∴ Point S is the intersection point of AB and PQ.

PQ : x + y + 1 = 0

AB : y - x = 6 or y = x + 6

Substituting value of y from equation AB in equation PQ, we get :

⇒ x + (x + 6) + 1 = 0

⇒ 2x + 7 = 0

⇒ 2x = -7

⇒ x = $-\dfrac{7}{2}$

Substituting value of x in equation AB, we get :

y = $-\dfrac{7}{2} + 6 = \dfrac{-7 + 12}{2} = \dfrac{5}{2}$.

S = $\Big(-\dfrac{7}{2}, \dfrac{5}{2}\Big)$.

Hence, coordinates of R = (-1, 0) and S = $\Big(-\dfrac{7}{2}, \dfrac{5}{2}\Big)$.