Question

A line is drawn from $P(x_1,y_1)$ in the direction $\theta$ with the X - axis, to meet $ax+by+c=0$ at $Q$. Then length $PQ$ is equal to :

• A. $\frac{|a{x}_1+b{y}_1+c|}{\sqrt{(a^2+b^2)}}$
• B. $\left|\frac{a{x}_1+b{y}_1+c}{acos\theta +bsin\theta }\right|$
• C. $\frac{b{x}_1+a{y}_1+c}{acos\theta +bsin\theta }$
• D. $-\frac{a{x}_1+b{y}_1+c}{asin\theta +bcos\theta }$

Answer 1

Answer: (B)

$\left|\frac{a{x}_1+b{y}_1+c}{acos\theta +bsin\theta }\right|$

Equation of a line drawn through a point $P\left(x_1{y}_1\right)$ at an angle $\theta$ with the $x-axis$

$\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }⟶\left(1\right)$

$Q$ is a point the above line and also lies on the line $ax+by+cz=0$

Say $\left|PQ\right|=r$ and the coordinates of $Q$ are $(h,k)$

Then,

$h=x_1+r\cos \theta$

$k=y_1+r\sin \theta$

$\because$ $Q$ lies on $ax+by+cz=0$

$\therefore$ $ah+bk+c=0$

$\Rightarrow a(x_1+r\cos \theta )+b(y_1+r\sin \theta )+c=0$

$\Rightarrow {ax}_1+{by}_1+c+r(a\cos \theta +b\sin \theta )=0$

$\Rightarrow r=\frac{-({ax}_1+{by}_1+c)}{a\cos \theta +b\sin \theta }$

$\because$ ${}^{\prime }{r}^{\prime }$ is the magnitude of length of line segment $PQ$, it cannot be negative.

So, $r=\left|\frac{{ax}_1+{by}_1+c}{a\cos \theta +b\sin \theta }\right|$