Question

Two straight lines are drawn through the point $(0,2)$ such that the length of the perpendiculars from the point $(4,4)$ to these lines are each equal to $2$ units. The equation of the line joining the feet of these perpendiculars is

• A. $y+x=5$
• B. $2y+3x=8$
• C. $y-2x=10$
• D. $y+2x=10$

Answer 1

Answer: (D)

Let equation of line passing through $(0,2)$ is

$y-2=mx$
$y=mx+2$
$BC=2$
$\therefore 2=\left|\frac{4m-4+2}{\sqrt{1+m^2}}\right|$
$\sqrt{1+m^2}=(2m-1)$
$\Rightarrow 1+m^2=4m^2-4m+1$
$\Rightarrow m=0,\frac{3}{4}$
$m=0,y=2$
$\therefore D(4,2)$
Equation of line passing through $D$ and $C$ is perpendicular to $AB$.
$\therefore$ Required equation of line
$y-2=\frac{-(4-0)}{4-2}(x-4)$
$y-2=-2(x-4)$
$y+2x=10$