ICSE Class 10 • 2024 • 4 Marks

Coordinate Geometry: Section Formula and Ratio of Division

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

Find the ratio in which the line segment joining the points $A(-3, 10)$ and $B(6, -8)$ is divided by the point $P(-1, 6)$. Also verify the coordinate using the $y$-value.

Verified Solution & Marking Scheme

Let ratio be k : 1 and set up x-coordinate relation
Let point $P(-1, 6)$ divide $AB$ in the ratio $k : 1$. Using internal section formula for $x$-coordinate: $x = \frac{k(x_2) + 1(x_1)}{k + 1} \implies -1 = \frac{k(6) + 1(-3)}{k + 1}$ $-(k + 1) = 6k - 3 \implies -k - 1 = 6k - 3$ $7k = 2 \implies k = \frac{2}{7}$
Verify with y-coordinate
Substitute $k = \frac{2}{7}$ into section formula for $y$: $y = \frac{k(y_2) + 1(y_1)}{k + 1} = \frac{\frac{2}{7}(-8) + 1(10)}{\frac{2}{7} + 1} = \frac{-\frac{16}{7} + \frac{70}{7}}{\frac{9}{7}} = \frac{54/7}{9/7} = \frac{54}{9} = 6$ This matches the given $y$-coordinate $6$.
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