Mathematics — CBSE Class 10 board question

Given: P(−2, 5) and Q(3, 2). R lies on PQ such that PR = 2QR.

$$\Rightarrow \frac{PR}{QR} = \frac{2}{1}$$

So R divides PQ internally in the ratio 2 : 1.

Here, $x_1 = -2,\ y_1 = 5,\ x_2 = 3,\ y_2 = 2,\ m_1 = 2,\ m_2 = 1$.

Using the Section Formula:

$$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{2(3) + 1(-2)}{2+1} = \frac{6-2}{3} = \frac{4}{3}$$

$$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{2(2) + 1(5)}{2+1} = \frac{4+5}{3} = 3$$

$$\therefore \text{Coordinates of R} = \left(\frac{4}{3},\ 3\right)$$

Source: Chapter 7, Section 7.3 (Section Formula)

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• The key step is converting PR = 2QR into a ratio: PR : QR = 2 : 1. Students often miss this and apply the formula incorrectly.
• R is between P and Q (internal division), so the Section Formula applies directly.
• Always identify which point is $(x_1, y_1)$ and which is $(x_2, y_2)$ — here P is the first point and Q is the second, matching the ratio $m_1 : m_2 = 2 : 1$ (PR corresponds to $m_1$, QR to $m_2$).
• Show the formula, substitution, and simplification clearly for full 3-mark credit.