Chapter 9.4, Problem 33E

To determine

To find: the polar form of the equations of two lines and neither of which is vertical and lines intersect at orthogonal angle.

Answer to Problem 33E

  $2=\cos \left(\theta -\left(\alpha +{90}^0\right)\right)$ .

Explanation of Solution

Given:

Two lines $l_1,l_2$ are perpendicular, and have normal segments of length $2$ units.

Concept used:

The polar form of a complex number is another way to represent a complex number. The form

$z=x+iy$ is rectangular coordinate and in polar from $x=r\cos \theta ,y=r\sin \theta$ and the polar form of equation:

  $z=r\left(\cos \theta +i\sin \theta \right)$ .

Calculation:

It is known that the polar equation of the line is:

  $p=r\cos \left(\theta -\varphi \right)\mathrm{.......}\left(i\right)$

Here $p$ is the length of the normal and $\varphi$ is the positive angle between the positive $x$ axis and the normal.

Let’s suppose that two lines $l_1,l_2$ are perpendicular, and have normal segments of length $2$ units.

If the angle between the positive $x$ -axis and the line is $\alpha$ , then the equation of the line $l_1$ can be taken as:

  $2=\cos \left(\theta -\alpha \right)$ .

Since, the line $l_1,l_2$ are perpendicular, the angle between the positive $x$ -axis and the line $l_2$ is $\alpha +{90}^0$ ,

then the equation of the line $l_2$ can be taken as:

  $2=\cos \left(\theta -\left(\alpha +{90}^0\right)\right)$ .