Question

Find the point which divides, internally and externally, the line joining $(-3,-4)$ to $(-8,7)$ in the ratio $7:5$.

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
$x=\frac{x_1\times n+x_2\times m}{m+n}$
$y=\frac{y_1\times n+y_2\times m}{m+n}$
If point
(i) divides the line internally, take $m$ & $n$ positive
(ii) divides the line externally, take any one of them negative
Given : $A(-3,-4),B(-8,7)$ and ratio $7:5$
(a) Let P$(x_1,y_1)$ divides internally
$x_1=\frac{(-3)\times 5+(-8)\times 7}{7+5}$
$\Rightarrow x_1=\frac{-71}{12}$

$y_1=\frac{(-4)\times 5+7\times 7}{7+5}$
$\Rightarrow y_1=\frac{29}{12}$
So, point P$(\frac{-71}{12},\frac{29}{12})$ divides internally
(b) Let Q$(x_2,y_2)$ divides externally
let us take $n$ negative i.e. $-5$, then becomes ratio $7:-5$
$x_2=\frac{(-3)\times (-5)+(-8)\times 7}{7+(-5)}$
$\Rightarrow x_2=\frac{-41}{2}$
$y_2=\frac{(-4)\times (-5)+7\times 7}{7+(-5)}$
$\Rightarrow y_2=\frac{69}{2}$
Thus, point $Q(\frac{-41}{2},\frac{69}{2})$ divides externally