Mathematics — Standard (041) — CBSE Class 10 board question

Let the y-axis divide the line segment joining A(5, –6) and B(–1, –4) in the ratio k : 1.

Since the point lies on the y-axis, its x-coordinate = 0.

Using the section formula:

$$x = \frac{k(-1) + 1(5)}{k + 1} = 0$$

$$\Rightarrow -k + 5 = 0 \Rightarrow k = 5$$

So the ratio is 5 : 1.

Now, the y-coordinate of the point of intersection:

$$y = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = \frac{-13}{3}$$

The point of intersection is $\left(0,\ \dfrac{-13}{3}\right)$.

Source: Chapter 7, Section 7.3 (Section Formula)

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• A point on the y-axis always has x-coordinate = 0. Use this condition to find k first.
• Apply the section formula for both x and y coordinates: $\left(\dfrac{m_1x_2 + m_2x_1}{m_1+m_2},\ \dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$.
• Examiners award marks for: correctly setting up k : 1, finding k = 5, and calculating the y-coordinate. Show all steps clearly.