Question

Find the ratio in which the line segment joining points (-3, 10) and (6, - 8) is divided by (-1, 6).

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio m:n, then

$(x,y)$ = $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let the ratio in which the line segment joining A(-3, 10) and B(6, - 8) is divided by

P(-1, 6) be k : 1.

Then, by section formula, the coordinates of P are $(\frac{6k-3}{k+1},\frac{-8k+10}{k+1})$

$⟹(-1,6)=(\frac{6k-3}{k+1},\frac{-8k+10}{k+1})$

Therefore, $-1=\frac{6k+1(-3)}{k+1}$
$\Rightarrow$ $-k-1=6k-3$
$\Rightarrow$ $2=7k$
$\Rightarrow$ $k=\frac{2}{7}$
$\Rightarrow$ $k:1=\frac{2}{7}:1=2:7$
Therefore, the point P divides AB in the ratio 2:7.