Question

The distance of the point $(2, 3)$ from the line $ 2\,x-3\,y+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 $2 x-3 y+9=0$,
then $4+\sqrt{2} r-9-\frac{3 r}{\sqrt{2}}+9=0$
$\Rightarrow r=4 \sqrt{2}$
Alternate Method

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