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Q) Find the ratio in which the x-axis divides the line segment joining the points(− 6, 5) and (− 4, − 1). Also, find the point of intersection.

(Q31 – 30/1/3 – CBSE 2026 Question Paper)

Ans: 

Let’s consider the points are given as: P (- 6, 5) and Q (- 4, – 1)

(i) Ratio of division:

Step 1: Let’s consider the coordinates of point A as (x, 0) and that the line PQ is 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:

(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n})

Here, we have following data: 

P (- 6, 5) = (x1, y1​)

Q (- 4, – 1) = (x2​, y2​),

Line PQ is divided in the ratio of m : n

Hence the y-coordinate of point A:

y = \frac{m (- 1) + n (5)}{(m + n)}

Step 2: We know that on X- axis, value of y-coordinate, is always 0

∵ y = 0, ∴  0 = \frac{m (- 1) + n (5)}{(m + n)}

∴ – m + 5 n = 0 

∴ m = 5 n

∴ m : n = 5 : 1

Therefore, the line is divided in the ratio of 5 : 1

(ii) Coordinates of intersection point:

Step 3: From the section formula, let’s find the value of x coordinate:

x = \frac{m x_2 + n x_1}{m + n}

∵ m : n = 5 : 1

∴ we can consider, m = 5 k and n = k, if k in an integer.

∴ x = \frac{(5 k) (- 4) + (k)(- 6)}{5 k + k}

∴ x = \frac {- 20 k - 6 k}{6 k} = \frac {- 26 k}{6 k}=\frac {- 13}{3}

Since, value of y coordinate is 0 (being on X-axis)

Therefore, the coordinates of intersection point A are (\frac {- 13}{3}, 0).

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