Q) P is a point on the x- axis which divides the line joining A (- 6, 2) and B (9, – 4). Find:

(a) the ratio in which P divides the line segment AB.

(b) the coordinates of the point P.

(c) equation of a line parallel to AB and passing through (-3, -2).

ICSE Specimen Question Paper (SQP) 2026

Ans: Let’s draw a tentative diagram for our better understanding:

(a) Ratio of division:

Step 1: It is given that X-axis intersects the line connecting points A & B

and we know that the value of y-coordinate of any point on X- axis, is always 0

hence, the value of y-coordinate of intersection point P will be 0.

∴ we can consider the coordinates of point P as (x, 0)

Step 2: Next, let’s consider that the line AB gets divided in the ratio of m : n.

We know that, by section formula, if a point (x, y) divides the line joining the points (x₁, y₁​) and (x₂​, y₂​) in the ratio m : n, then the coordinates of intersection point (x, y) is given by:

Here, it is given that

A (- 6, 2) = (x₁, y₁​)

B (9, – 4) = (x₂, y₂​)

Step 3: Let’s consider line is divided in the ratio of m : n

Hence, the y-coordinate of point A:

y =

∴  0 =

∴ – 4 m + 2 n = 0

∴ 4 m = 2 n

∴

∴ m : n = 1 : 2

Therefore, the line is divided in the ratio of 1 : 2.

(b) Coordinates of intersection point:

We considered coordinates of point P are (x, 0)   (calcualted in step 1 above)

Step 4: From the section formula, let’s find the value of x coordinate:

x =

Given points are: A (- 6, 2) and B (9, – 4) and m : n = 1 : 2 (calculated above)

∴ x =

∴ x = = 3

∴ x = = – 1

Therefore, the coordinates of intersection point P are (- 1, 0).

(c) Line parallel to AB, passing through (-3, -2):

Given points are: A (- 6, 2) and B (9, – 4)

Step 5: Let’s first calculate gradient or slope of the line AB

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

m =

∴ slope of line AB, m_AB passing through points A (- 6, 2) and B (9, – 4) will be:

m_AB =  =

=  =

Step 6: Now the line parallel to AB will have same slope i.e. its m =

(Extra Notes: The slope of parallel lines has to be equal. If slopes are not same, 2 lines will definitely intersect each other)

Step 7:  We know that, the equation of the line parallel to AB and passing through point (x₁, y₁) is given by:

y – y₁ = m (x – x₁)

We are given that the new line is passing through the point (-3, -2)

∴ y – (-2) = ( x – (-3)

∴ y + 2 = ( x + 3)

∴ 5(y + 2) = (- 2) (x + 3)

∴ 5y  + 10 = – 2 x – 6

∴ 2x + 5y + 10 + 6 = 0

∴ 2x + 5y + 16 = 0

Therefore, the equation of the line is 2x + 5y + 16 = 0

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