Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Straight Lines
0 votes

Reduce the following equation into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive $x$ - axis. $ x-\sqrt 3 y +8=0$

$\begin{array}{1 1}(A)\;p = 4\: units\: and \: \alpha = 120^{\circ} \\(B)\; p = 4\: units\: and \: \alpha = 60^{\circ} \\(C)\; p = 8\: units\: and \: \alpha = 120^{\circ} \\(D)\;p = 8\: units\: and \: \alpha = 60^{\circ} \end{array} $

Can you answer this question?

1 Answer

0 votes
  • Equation of a line in the normal form is $ x \cos \alpha + y \sin \alpha=p$ where p is the perpendicular distance from the origin and $ \alpha$ is the angle between the perpendicular and the positive $x$ - axis.
Given equation is $ x-\sqrt 3y=-8$
This can be written as
$ \qquad x-\sqrt 3y=-8$
or $ \quad -x+\sqrt 3y=8$
$ \sqrt{(-1)^2+(\sqrt 3)^2}=\sqrt 4 = 2$
dividing by 2 on both sides we get,
$\qquad -\large\frac{x}{2}$$+\large\frac{\sqrt 3}{2}$$y=4$
$ \Rightarrow \bigg( -\large\frac{1}{2}$$\bigg)(x)+ \bigg( \large\frac{\sqrt 3}{2}$$\bigg)(y)=4$
This implies that $\cos \alpha = -\large\frac{1}{2}$ and $\sin \alpha = \large\frac{\sqrt 3}{2}$.
That is the angle is in the second quadrant .
$ \therefore \alpha = 120^{\circ}$
$ \Rightarrow x \cos 120^{\circ}+y \sin 120^{\circ}=4$
which is in the normal form.
Comparing the above equation with the normal form of the equation we get, $x \cos \alpha + y \sin \alpha=p$
Hence $p=4$
That is the perpendicular distance from the origin to the line is 4 units.
The angle $ \alpha = 120^{\circ}$
answered May 7, 2014 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App