Determine the ratio in which $y - x + 2 = 0$ divides the line segment joining $(3, - 1)$ and $(8, 9)$ .

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Solution

Given the equation of line $y - x + 2 = 2$

Let the points be denoted as $A (3, - 1) \equiv (x_{2}, y_{2})$ and $B (8, 9) \equiv (x_{1}, y_{1})$

Let the line segments joining the points $A$ and $B$ be of ratio
$K : 1 a t p o i n t C$ .

Here $m = K$ and $n = 1$

By section formula,

$[x = (\frac{m x_{1} + n x_{2}}{m + n})]$ and $[y = (\frac{m y_{1} + n y_{2}}{m + n})]$

$∴ C o o r d i n a t e s o f C = (\frac{8 K + 3}{K + 1}, \frac{9 K - 1}{K + 1})$

$C [\frac{8 K + 3}{K + 1}, \frac{9 K - 1}{K + 1}]$ lies on the line $y - x + 2 = 0$

$\Rightarrow \frac{9 K - 1}{K + 1} - \frac{8 K + 3}{K + 1} + 2 = 0$
$\Rightarrow 9 K - 1 - 8 K - 3 + 2 K + 2 = 0$

$3 K - 2 = 0 \Rightarrow K = \frac{2}{3}$

$∴$ The ratio $= 2 : 3$

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