A line segment joining P(2,- 3) and Q (0, -1) is cut by the x-axis

Q) 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.

ICSE Specimen Question Paper (SQP)2025

Ans:

(a) Equation of line PQ:

Step 1: We know that the line passing through points (x₁, y₁) and (x₂, y₂) is given by:

(y – y_1) = $\frac{y_2 - y_1}{x_2 - x_1}$ (x – x_1)

∴ PQ line passing through P (2, – 3) and Q (0, – 1) is given by:

(y – (- 3)) = $\frac{(- 1) - (- 3)}{0 - 2}$ (x – 2)

∴ (y + 3) = $\frac{2}{- 2}$ (x – 2)

∴ y + 3 = ( – 1) ( x – 2)

∴ y + 3 = – x + 2

∴ x + y + 1 = 0

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

(b) Equation of line AB:

Step 2: Let’s first calculate gradient or slope of the line PQ

We know that the slope of a line passing through points (x₁, y₁) and (x₂, y₂) is given by:

m = $\frac{y_2 - y_1}{x_2 - x_1}$

∴ Slope of line PQ, m_PQ passing through points P (2,- 3) and Q (0, – 1) will be:

m_PQ = $\frac{(- 1) - (- 3)}{(0) - (2)} = \frac{(- 1 + 3)}{(0 - 2)}$

∴ m_PQ = $\frac{2}{- 2}$ = – 1

Step 3: Next, we calculate the slope of line AB.

We know that the slopes of perpendicular lines are negative reciprocals of each other. Therefore, the product of the slopes of two perpendicular lines is -1.

∴ m_AB x m_PQ = – 1

∴ m_AB x (- 1) = – 1

∴ m_AB = 1

Step 4: The equation of the line of slope m and passing through point (x₁, y₁) is given by:

y – y₁= m (x – x₁)

∴ Equation of line AB, with slope (m_AB = 1) and passing through point T (0, 6) is:

y – 6= (1) (x – 0)

y – 6 = x

x – y + 6 = 0

Therefore, equation of line AB is x – y + 6 = 0

(b) Coordinates of points R and S:

Step 5: Given that the line PQ cuts x – axis at point R, hence its ordinate value is 0 i.e. y = 0

Substituting this value in PQ, we get:

x + y + 1 = 0

x + (0) + 1 = 0

x = – 1

Therefore, the coordinates of point R are: (- 1, 0).

Step 6: Since line PQ and AB intersect each other at point S, hence S will satisfy both the equations. We can get the coordinates by solving these equations for values of x and y.

PQ: x + y + 1 = 0

∴ x + y = – 1 ……….. (i)

AB: x – y + 6 = 0

∴ x – y = – 6 ………… (ii)

By adding equations (i) and (ii), we get:

(x + y) + (x – y) = (- 1) + (- 6)

2 x = – 7

x = $\frac{-7}{2}$