Question

Two mutually perpendicular straight lines are drawn from the origin to form an isosceles triangle with the straight line $x\cos \alpha +y\sin \alpha -p=0$ .Then the area of this triangle is
(a) Independent of $\alpha$ .
(b) Independent of p .
(c) Independent of both $\alpha$ and p .
(d) A function of $\alpha$ and p .

Answer 1

Hint: Here, we have been given the normal form of the equation of the line. Using the given equation, we can find the length of the perpendicular from the origin to the given line. Then, we will find the length of the perpendicular and the base of the right angled triangle and then we can find its area.

Complete step-by-step answer:

The equation of the straight line upon which the length of the perpendicular from the origin is p and this perpendicular makes an angle $\alpha$ with x-axis is :
$x\cos \alpha +y\sin \alpha =p$

Let us suppose that the line cuts the x-axis at A and the y-axis at B. We can draw a perpendicular from the origin to the line which cuts the line at D.
The length of the perpendicular from the origin = p
And, the angle made by the perpendicular with the x-axis is = $\alpha$
Thus, we can write:
$\begin{aligned} & \cos \alpha =\dfrac{\text{length of perpendicular from the origin}}{\text{x-intercept of the line}} \\\ & \Rightarrow \cos \alpha =\dfrac{p}{OA} \\\ & \Rightarrow OA=\dfrac{p}{\cos \alpha } \\\ \end{aligned}$
It means that the x-intercept of the line is $\dfrac{p}{\cos \alpha }$.

Again, we can write:
$\begin{aligned} & \sin \alpha =\dfrac{\text{length of perpendicular from the origin}}{\text{y-intercept of the line}} \\\ & \Rightarrow \sin \alpha =\dfrac{p}{OB} \\\ & \Rightarrow OB=\dfrac{p}{\sin \alpha } \\\ \end{aligned}$
It means that the y-intercept of the line is $\dfrac{p}{\sin \alpha }$ .
Since, the triangle so formed is isosceles. So, we have OA=OB
Therefore,
$\begin{aligned} & \dfrac{p}{\cos \alpha }=\dfrac{p}{\sin \alpha } \\\ & \Rightarrow \dfrac{\sin \alpha }{\cos \alpha }=\dfrac{p}{p} \\\ & \Rightarrow \tan \alpha =1 \\\ & \Rightarrow \alpha =\dfrac{\pi }{4} \\\ \end{aligned}$
So, the length of OA or x-intercept of the line is =$\dfrac{p}{\cos \dfrac{\pi }{4}}=\dfrac{p}{\dfrac{1}{\sqrt{2}}}=\sqrt{2}p$

And, length of OB or y-intercept of the line is =$\dfrac{p}{\sin \dfrac{\pi }{4}}=\dfrac{p}{\dfrac{1}{\sqrt{2}}}=\sqrt{2}p$
So, the area of the triangle will be:
$\begin{aligned} & =\dfrac{1}{2}\times base\times height \\\ & =\dfrac{1}{2}\times OA\times OB \\\ & =\dfrac{1}{2}\times \sqrt{2}p\times \sqrt{2}p \\\ & =\dfrac{2{{p}^{2}}}{2}={{p}^{2}} \\\ \end{aligned}$

It shows that the area of the triangle is independent of $\alpha$.
Hence, option (a) is the correct answer.

Note: Students should note here that we are equating lengths of x-intercept and y-intercept because it is given that the triangle so formed is isosceles. So, we used the concept that in an isosceles triangle, the lengths of two sides are equal.