No products in the cart.

Find the equation of a line …

CBSE, JEE, NEET, CUET

CBSE, JEE, NEET, CUET

Question Bank, Mock Tests, Exam Papers

NCERT Solutions, Sample Papers, Notes, Videos

Find the equation of a line whose perpendicular distance from the Origin is 4 unit and the angle which the normal makes with positive direction of x axis is 15°
  • 1 answers

Preeti Dabral 1 year, 8 months ago

We are given that, p = 4 and ω = 150
Now, {tex}\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}{/tex}
and {tex}\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}{/tex}
The equation of the line is x cos ω + y sin ω = p
{tex}x \cos 15^{\circ}+y \sin 15^{\circ}{/tex}
or {tex}\frac{\sqrt{3}+1}{2 \sqrt{2}} x+\frac{\sqrt{3}-1}{2 \sqrt{2}} y=4{/tex} 
or {tex}(\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}{/tex}
This is the required equation.

http://mycbseguide.com/examin8/

Related Questions

2nC2:nC3=33:10
  • 0 answers
Square of 169
  • 1 answers
Ch 1 ke questions
  • 1 answers
Express the complex number i-39
  • 0 answers
(3+i)x + (1-2i) y +7i =0
  • 1 answers

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App