Question

The distance of the point $(2,3)$ from the line $2x-3y+9=0$ measured along a line $x-y+1=0$ is

• A. $4\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A)

The slope of the line $x-y+1=0$ is $1.$
So, it makes an angle of $45^o$ with $x-$ axis.
The equation of a line passing through $(2,3)$ and making an angle of $45^{\circ }$
$\frac{x-2}{\cos 45^{\circ }}=\frac{y-3}{\sin 45^{\circ }}=r$
Coordinates of any point on this line are
$\left(2+r\cos 45^{\circ },3+r\sin 45^{\circ }\right)$
or $\left(2+\frac{r}{\sqrt{2}},3+\frac{r}{\sqrt{2}}\right)$
If this point lies on the line $2x-3y+9=0$,
then $4+\sqrt{2}r-9-\frac{3r}{\sqrt{2}}+9=0$
$\Rightarrow r=4\sqrt{2}$
Alternate Method

Since thepoint $(2,3)$ lies on the line $x-y+1=0$. Therefore the distance from $(2,3)$ to the line $2x-3y+9=0$ along the line $x-y+1=0$ is equal to the distance between the points $(2,3)$ and intersection point of $2x-3y+9=0$ and $x-y+1=0$ ie, (6, 7).
Hence required distance
$d=\sqrt{(6-2{)}^2+(7-3{)}^2}=\sqrt{32}$
$d=4\sqrt{2}$