Question

Find the distance of the line $4x+7y+5$ $=$ $0$ from the point $(1,2)$ along the line $2x-y=$ $0$.

Answer 1

The given lines are
$2x-y=0$...(1)
$4x+7y+5=0$....(2)
$A(1,2)$ is point on line (1).
Let $B$ be the point of intersection of lines (1) and (2).
On solving equation (1) and (2), we obtain$x=\frac{-5}{18}\text{and}y=\frac{-5}{9}$
$\therefore$coordinates of point $B$ are$\{\frac{-5}{18},\frac{-5}{9}\}$
By using distance formula, the distance between point $A$ and $B$ can be obtained as$AB=\sqrt{{\{1+\frac{5}{18}\}}^2+{\{2+\frac{5}{9}\}}^2}units$
$=\sqrt{{\left\{\frac{23}{18}\right\}}^2+{\left\{\frac{23}{9}\right\}}^2}units$
$=\sqrt{{\left\{\frac{23}{2\cdot 9}\right\}}^2+{\left\{\frac{23}{9}\right\}}^2}units$
$=\sqrt{{\left\{\frac{23}{9}\right\}}^2{\left\{\frac{1}{2}\right\}}^2+{\left\{\frac{23}{9}\right\}}^2}units$
$=\sqrt{{\left(\frac{23}{9}\right)}^2(\frac{1}{4}+1)}units$
$=\frac{23}{9}\sqrt{\frac{5}{4}}units$
$=\frac{23}{9}\cdot \frac{\sqrt{5}}{2}units$
$=\frac{23\sqrt{5}}{18}units$
Thus, the required distance is$=\frac{23\sqrt{5}}{18}units$