Question

The distance of origin to the line $3x+4y-5=0$ measured along the line $x-y+2=0$ is

• A. $1\text{units}$
• B. $\sqrt{2}\text{units}$
• C. $\frac{\sqrt{2}}{7}\text{units}$
• D. $\frac{5\sqrt{2}}{7}\text{units}$

Answer 1

The correct option is D $\frac{5\sqrt{2}}{7}\text{units}$

Let $l_1:3x+4y-5=0,l_2:x-y+2=0$
Slope of line $l_2$ is
$m=1\Rightarrow \theta =\frac{\pi }{4}$
Now, equation of line passing through origin and having slope $1$ is
$\frac{x-0}{\cos \frac{\pi }{4}}=\frac{y-0}{\sin \frac{\pi }{4}}=r$
Any point on this line is
$(r\cos \frac{\pi }{4},r\sin \frac{\pi }{4})$
This point lies on $l_1$, we get
$3r\cos \frac{\pi }{4}+4r\sin \frac{\pi }{4}-5=0\Rightarrow r=\frac{5\sqrt{2}}{7}$

Hence, the required distance
$=\frac{5\sqrt{2}}{7}\text{units}$