Question

Find distance of point $A(2,3)$ measured parallel to the line $x-y=5$ from the line $2x+y+6=0$.

• A. $\frac{13\sqrt{2}}{3}$ units
• B. $\frac{13}{3}$ units
• C. $\frac{13\sqrt{2}}{6}$ units
• D. None of these

Answer 1

Answer: (A)

$\frac{13\sqrt{2}}{3}$ units

Consider the line passing through $(2,3)$ and parallel to $x-y=5$ to $x-y=c$
$c=x-y=>c=2-3=-1$
Now this intersects line $2x+y+6=0$ at some point and distance between this point and $(2,3)$ is the required distance.
Solving for $x-y+1=0$ and $2x+y+6=0$
$3x+7=0=>x=\frac{-7}{3}$ and $y=\frac{-4}{3}$
Distance between $(2,3)and(\frac{-7}{3},\frac{-4}{3})is\sqrt{(2+\frac{7}{3}{)}^2+(3+\frac{4}{3}{)}^2}=\sqrt{(\frac{13}{3}{)}^2+(\frac{13}{3}{)}^2}=\frac{13\sqrt{2}}{3}$