Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{L}_1$ is a line passing through the point $\mathrm{P}(4,-3)$ and perpendicular to the line $3 x-4 y+k=0$, then the distance of $P$ from the line $5 x-3 y-2=0$ measured along the line $\mathrm{L}_1$ is
Options:
Solution:
2174 Upvotes
Verified Answer
The correct answer is:
$5$
Slope of line perpendicular to $3 x-4 y+k=0$ is $\frac{-4}{3}$
$$
\therefore L_1: y+3=\frac{-4}{3}(x-4) \Rightarrow 3 y+9=-4 x+16
$$
$L_1: 4 x+3 y=7$ and $L: 5 x-3 y=2$ (given)
Point of intersection of $L_1$ and $L:(x, y) \equiv(1,1)$
$\therefore \quad$ Distance between $P(4,-3)$ and $(1,1)$
$$
=\sqrt{(4-1)^2+(-3-1)^2}=5 \text {. }
$$
$$
\therefore L_1: y+3=\frac{-4}{3}(x-4) \Rightarrow 3 y+9=-4 x+16
$$
$L_1: 4 x+3 y=7$ and $L: 5 x-3 y=2$ (given)
Point of intersection of $L_1$ and $L:(x, y) \equiv(1,1)$
$\therefore \quad$ Distance between $P(4,-3)$ and $(1,1)$
$$
=\sqrt{(4-1)^2+(-3-1)^2}=5 \text {. }
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.