Question

Find the equation of the straight line perpendicular to $5x-2y=8$ and which passes through the mid-point of the line segement joining $(2,3)$ and $(4,5)$.

Answer 1

The equation of required line is

$y-y_1=m(x-x_1)...\left(1\right)$

$\left(x_1{y}_1\right)$ is mid point of $\begin{array}{c}(2,3)\\ \left(x_1{y}_1\right)\end{array}$ and $\begin{array}{c}(4,5)\\ \left(x_2{y}_2\right)\end{array}$

$\Rightarrow x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

$\Rightarrow x=\left(\frac{2+4}{2}\right),y=\left(\frac{3+5}{2}\right)$

$\Rightarrow \left(x_1{y}_1\right)\iff (3,4)$

Also slope of given line is $5x-2y=8$

$y=\frac{5}{2}x-4$

$\Rightarrow m^{\prime }=\frac{5}{2}$

Required line is perpendicular to the given line.

$\therefore m=\frac{-2}{5}$

Putting m and $(x_1,y_1)$ in (1)

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

$5y+2x=26$

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