Question

The distance of the point $(1,2)$ on the straight line with slope $5$ and passing through the point of intersection lines $x+2y=5$ and $x-3y=7$ is

• A. $\frac{64}{\sqrt{650}}$
• B. $\frac{16}{\sqrt{650}}$
• C. $\frac{102}{\sqrt{650}}$
• D. $5$

Answer 1

The correct option is C $\frac{102}{\sqrt{650}}$

The point of intersection of two lines is given by

$x+2y=5$ and $x-3y=7$

$\Rightarrow 5-2y-3y=7$

$y=\frac{-2}{5}$ and $x=\frac{29}{5}$

$\therefore$ equation of line having slope $5$ and passes through $(\frac{29}{5},\frac{-2}{5})$ is

$(y+\frac{2}{5})=5(x-\frac{29}{5})$

$y=5x-29-\frac{2}{5}$

$y=5x-\frac{147}{5}$

So, perpendicular distance from $(1,2)$ to $y=5x-\frac{147}{5}$

$d=\left|\frac{1-10+\frac{147}{5}}{\sqrt{26}}\right|$

$=\left|\frac{102}{5\sqrt{26}}\right|$

$d=\left|\frac{102}{\sqrt{650}}\right|$