Straight Lines
5.0 Length of perpendicular from a point to a line
5.0 Length of perpendicular from a point to a line
The length of perpendicular from a point $P({x_1},{y_1})$ to the line $ax+by+c=0$ is
$$\left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$$
Question 5. If $p$ and $p'$ are the length of the perpendiculars from the origin to the straight lines whose equations are $x\sec \theta + y\operatorname{cosec} \theta = a$ and $x\cos \theta - y\sin \theta = a\cos 2\theta $, then find the value of $4{p^2} + p{'^2}$.
Solution: We have $$p = \frac{{\left| { - a} \right|}}{{\sqrt {{{\sec }^2}\theta + {{\operatorname{cosec} }^2}\theta } }}$$
Squaring both sides, we get $$\begin{equation} \begin{aligned} {p^2} = \frac{{{a^2}}}{{{{\sec }^2}\theta + {{\operatorname{cosec} }^2}\theta }} = \frac{{{a^2}{{\sin }^2}\theta {{\cos }^2}\theta }}{1} \\ 4{p^2} = {a^2}{\sin ^2}2\theta \\\end{aligned} \end{equation} $$
and $$p' = \frac{{\left| { - a\cos 2\theta } \right|}}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }} = \left| { - a\cos 2\theta } \right|$$
Squaring both sides, we get $$p{'^2} = - {a^2}{\cos ^2}2\theta $$
Now, put the values in $$4{p^2} + p{'^2}$$We get $${a^2}{\sin ^2}2\theta + {a^2}{\cos ^2}2\theta = {a^2}$$
