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

Let P(x, 7) divide AB in the ratio k : 1, where A(−5, 4) and B(7, 9).

Using the section formula for the y-coordinate:

$$7 = \frac{k \times 9 + 1 \times 4}{k + 1}$$

$$7(k + 1) = 9k + 4$$

$$7k + 7 = 9k + 4$$

$$2k = 3 \implies k = \frac{3}{2}$$

So the ratio is 3 : 2.

Now, using the section formula for the x-coordinate:

$$x = \frac{k \times 7 + 1 \times (-5)}{k + 1} = \frac{\frac{3}{2} \times 7 + (-5)}{\frac{3}{2} + 1} = \frac{\frac{21}{2} - 5}{\frac{5}{2}} = \frac{\frac{11}{2}}{\frac{5}{2}} = \frac{11}{5}$$

Therefore, the ratio is 3 : 2 and $x = \dfrac{11}{5}$.

Source: Chapter 7, Section 7.3 — Section Formula

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• The standard technique is to let the ratio be k : 1 (not m : n), which keeps the algebra to one unknown.
• Use the y-coordinate equation first (since y = 7 is given) to find k, then substitute k into the x-coordinate formula.
• Examiners award marks for: setting up the section formula correctly (1 mark), finding the ratio (1 mark), and finding x (1 mark).
• Write the final ratio as integers — simplify k = 3/2 to ratio 3 : 2.