Question

Find the equation of the straight line segment whose end points are the points of intersection of the straight lines $2x-3y+4=0,x-2y+3=0$ and the midpoint of the line joining the points $(3,-2)$ and $(-5,8)$.

Answer 1

The given lines are $2x-3y+4=0$ ... (1)
and $x-2y+3=0$ .... (2)
$2x-3y+4=0$ ..... (1)
$2$ $\times (2x-4y+6=0)$
(-) (+) (-)
------------------------------------------
$y-2=0$
$\Rightarrow y=2$
Substitute $y=2$ in equation (1), we get
$2x-6+4=0$
$\Rightarrow 2x-2=0$
$\Rightarrow 2x=2$
The point of intersection is $(1,2)$.
Mid point of line joining the points $(3,-2),(-5,8)$.
$=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{3-5}{2},\frac{-2+8}{2})=(-1,3)$
The required line is
$=\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
$=\frac{y-2}{3-2}=\frac{x-1}{-1-1}$

$\Rightarrow \frac{y-2}{1}=\frac{x-1}{-2}$
$=x-1=-24+4$

$\Rightarrow x+2y-5=0$