# Q : 15     Find the distance of the line  $\small 4x+7y+5=0$  from the point  $\small (1,2)$  along the line $\small 2x-y=0$.

point $\small (1,2)$ lies on line $2x-y =0$
Now, point of intersection of lines  $2x-y =0$    and    $\small 4x+7y+5=0$  is $\left ( -\frac{5}{18},-\frac{5}{9} \right )$
Now, we know that the distance between two point is given by
$d = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
$d = |\sqrt{(1+\frac{5}{18})^2+(2+\frac{5}{9})^2}|$
$d = |\sqrt{(\frac{23}{18})^2+(\frac{23}{9})^2}|$
$d = \left | \sqrt{\frac{529}{324}+\frac{529}{81}} \right |$
$d = \left | \sqrt{\frac{529+2116}{324}} \right | = \left | \sqrt\frac{2645}{324} \right | =\frac{23\sqrt5}{18}$
Therefore, the distance of the line  $\small 4x+7y+5=0$  from the point  $\small (1,2)$  along the line $\small 2x-y=0$  is   $\frac{23\sqrt5}{18} \ units$

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