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

Let the line $y = x$ divide the segment joining $A(3, -5)$ and $B(-1, 6)$ in the ratio $k:1$ at point $P$.

By the section formula:
$$x = \frac{k(-1) + 1(3)}{k+1} = \frac{3-k}{k+1}, \quad y = \frac{k(6) + 1(-5)}{k+1} = \frac{6k-5}{k+1}$$

Since $P$ lies on $y = x$:
$$\frac{3-k}{k+1} = \frac{6k-5}{k+1}$$
$$3 - k = 6k - 5 \implies 8 = 7k \implies k = \frac{8}{7}$$

∴ The ratio is $\mathbf{8:7}$.

Source: Chapter 7, Section 7.3 (Section Formula)

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• The key idea: any point on $y = x$ satisfies $x\text{-coordinate} = y\text{-coordinate}$.
• Use the section formula with ratio $k:1$, then set the two coordinates equal and solve for $k$.
• Examiners expect the section formula to be written out clearly, the condition $x = y$ applied, and the final ratio stated explicitly. No working step should be skipped for a 2-mark answer.