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

Finding coordinates of P (divides A(–1, 7) and B(4, –3) in ratio 3 : 2):

Using section formula with $m_1 = 3,\ m_2 = 2$:

$$x = \frac{3(4) + 2(-1)}{3+2} = \frac{12-2}{5} = \frac{10}{5} = 2$$

$$y = \frac{3(-3) + 2(7)}{3+2} = \frac{-9+14}{5} = \frac{5}{5} = 1$$

So P = (2, 1).

Verification: Substituting in $x - 3y = -1$:

$2 - 3(1) = 2 - 3 = -1$ ✓ Hence P lies on the line.

Length PA:

$$PA = \sqrt{(2-(-1))^2 + (1-7)^2} = \sqrt{9+36} = \sqrt{45} = 3\sqrt{5} \text{ units}$$

Length PB:

$$PB = \sqrt{(2-4)^2 + (1-(-3))^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5} \text{ units}$$

Source: Chapter 7, Section 7.3 (Section Formula) and Section 7.2 (Distance Formula)

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• Section formula is the core tool here: always substitute $m_1, m_2$ carefully and match them to the correct points.
• To prove P lies on the line, substitute the coordinates into the equation and show LHS = RHS.
• For distances PA and PB, apply the distance formula directly. Note that $PA : PB = 3\sqrt{5} : 2\sqrt{5} = 3 : 2$, which confirms the answer.
• Examiners award marks for: correct P coordinates (1 mark), verification on the line (1 mark), and both distances (1 mark).