Question

The distance of the point $(1,3)$ from the line $2x+3y=6$, measured parallel to the line $4x+y=4$, is

• A. $\frac{5}{\sqrt{13}}$units
• B. $\frac{3}{\sqrt{17}}$units
• C. $\sqrt{17}$units
• D. $\frac{\sqrt{17}}{2}$ units

Answer 1

The correct option is D $\frac{\sqrt{17}}{2}$ units

Equation of line L is
$y-3=-4(x-1)\Rightarrow y-3=-4x+4\Rightarrow 4x+y=7$

Solving equations, $x=\frac{3}{2},y=1$

$\therefore AP=\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}$

Therefore distance is $\frac{\sqrt{17}}{2}$