Question

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

• A. $x+2y-5=0$
• B. $x+2y+5=0$
• C. $x-2y-5=0$
• D. None of these

Answer 1

The correct option is A $x+2y-5=0$

The given lines are

$2x-3y+4=0$ $....\left(1\right)$

And

$x-2y+3=0$ $........\left(2\right)$

$\left(1\right)-\left(2\right)\times 2$

$2x-3y+4=0$

$2x-4y+6=0$

$\overline{0+y-2=0}$

$y=2$ substitute in $\left(1\right)$

$x-2\left(2\right)+3=0$

$x-1=0$

$x=1$

The point of intersection is

$(1,2)$

Mid point of line joining the points $(3,-2)$ and $(-5,8)$

$=(\frac{3-5}{2},\frac{-2+8}{2})$

$=(\frac{-2}{2},\frac{6}{2})=(-1,3)$

So, the required line is

$y-2=\frac{3-2}{-1-1}(x-1)$

$y-2=\frac{1}{-2}(x-1)$

$-2y+4=(x-1)$

$x+2y-5=0$

Hence, this is the answer.