Question

If the distance of the point $(2,3)$ from the line $2x-3y+9=0$ measured along a line $x-y+1=0$ is $k$ units, then the value of $k^2$ is

• A. 32
• B. 32.00
• C. 32.0

Answer 1

Answer: (A), (B), (C)

The slope of the line $x-y+1=0$ is $1$. So, it makes an angle of ${45}^{\circ }$ with $x-$axis.
Equation of the line passing through $(2,3)$ and making an angle of ${45}^{\circ }$ is
$\frac{x-2}{\cos {45}^{\circ }}=\frac{y-3}{\sin {45}^{\circ }}=r$
Coordinates of any point on this line are
$(2+r\cos {45}^{\circ },3+r\sin {45}^{\circ })\equiv (2+\frac{r}{\sqrt{2}},3+\frac{r}{\sqrt{2}})$
If this point lies on the line $2x-3y+9=0$, then
$4+\sqrt{2}r-9-\frac{3r}{\sqrt{2}}+9=0$
$\Rightarrow r=4\sqrt{2}\Rightarrow k=4\sqrt{2}\therefore k^2=32$