Question

Find the point which divides internally and externally, the line joining $(-1,2)$ to $(4,-5)$ in the ratio $2:3$.

Answer 1

Section formula :
Any point let say $(x,y)$ divides the line joining points $(x_1,y_1)$ & $(x_2,y_2)$ in the ratio $m:n$, then co-ordinates were given by the formula
$x=\frac{x_1\times n+x_2\times m}{m+n}$

$y=\frac{y_1\times n+y_2\times m}{m+n}$

Given points are $(-1,2)$ and $(4,-5)$ & ratio $2:3$

If (i) Points divides internally, take $m$ & $n$ positive

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

$y=\frac{2\times 3+(-5)\times 2}{5}=\frac{-4}{5}$

Then, $(x,y)=(1,\frac{-4}{5})$

(ii) Point divides externally, take any one of them negative

Take $m$ negative, the ratio becomes $-2:3$

$x=\frac{(-1)\times 3+4\times (-2)}{-2+3}=-11$

$y=\frac{2\times 3+(-5)\times (-2)}{-2+3}=16$

Then, $(x,y)=(-11,16)$.