Equation of a straight line on which length of the perpendicular from the origin is $4$ units and the line makes an angle of \[{120^ \circ }\] with the x-axis, is
(A) $x\sqrt 3 + y + 8 = 0$
(B) $x\sqrt 3 + y = 8$
(C) $x\sqrt 3 - y = 8$
(D) $x - \sqrt 3 y + 8 = 0$
Answer
300.9k+ views
Hint: In order to solve this question, first we will make the appropriate figure according to the question. Then, we will find the required angles using the linear pair property and angle sum property of the triangle. Now, to get the required equation of a line, we will use the above values obtained in the general equation of a line.
Complete step by step Solution:

Given,
\[\angle MAX = {120^ \circ }\] ………………..equation $\left( 1 \right)$
First, we will find $\angle MAO,$
$\angle MAO + \angle MAX = {180^ \circ }$ $[\because $ Linear Pair$]$
Put the value of \[\angle MAX\] and solve it,
\[\angle MAO + {120^ \circ } = {180^ \circ }\] $[\because $ Using equation $\left( 1 \right)]$
\[\angle MAO = {180^ \circ } - {120^ \circ }\]
\[\angle MAO = {60^ \circ }\] ………………..equation $\left( 2 \right)$
Now, let us find $\theta ,$
In $\vartriangle MOA,$
$\angle MOA + \angle MAO + \angle OMA = {180^ \circ }$ $[\because $ Angle Sum Property of Triangle$]$
\[\theta + {60^ \circ } + {90^ \circ } = {180^ \circ }\] $[\because $ Using given an equation $\left( 2 \right)]$
Solving it further,
$\theta = {30^ \circ }$ ………………..equation $\left( 3 \right)$
Now, let us find the required equation of the line,
$x\;\cos \theta + y\;\sin \theta = p$
Substituting values from equation $\left( 3 \right)$
$x\;\cos {30^ \circ } + y\;\sin {30^ \circ } = 4$ $\left[ {\because p = OM = 4} \right]$
$x\left( {\dfrac{{\sqrt 3 }}{2}} \right) + y\left( {\dfrac{1}{2}} \right) = 4$
Simplifying it further,
$x\sqrt 3 + y + 8 = 0$
This is the required equation of a line.
Hence, the correct option is (A).
Note:The key concept to solve this type of question is to have a basic knowledge of properties (Angle sum property of triangle, Linear pair property) learned in previous classes. The figure should be made with proper attention and labeled correctly. The value of trigonometric angles should be known.
Complete step by step Solution:

Given,
\[\angle MAX = {120^ \circ }\] ………………..equation $\left( 1 \right)$
First, we will find $\angle MAO,$
$\angle MAO + \angle MAX = {180^ \circ }$ $[\because $ Linear Pair$]$
Put the value of \[\angle MAX\] and solve it,
\[\angle MAO + {120^ \circ } = {180^ \circ }\] $[\because $ Using equation $\left( 1 \right)]$
\[\angle MAO = {180^ \circ } - {120^ \circ }\]
\[\angle MAO = {60^ \circ }\] ………………..equation $\left( 2 \right)$
Now, let us find $\theta ,$
In $\vartriangle MOA,$
$\angle MOA + \angle MAO + \angle OMA = {180^ \circ }$ $[\because $ Angle Sum Property of Triangle$]$
\[\theta + {60^ \circ } + {90^ \circ } = {180^ \circ }\] $[\because $ Using given an equation $\left( 2 \right)]$
Solving it further,
$\theta = {30^ \circ }$ ………………..equation $\left( 3 \right)$
Now, let us find the required equation of the line,
$x\;\cos \theta + y\;\sin \theta = p$
Substituting values from equation $\left( 3 \right)$
$x\;\cos {30^ \circ } + y\;\sin {30^ \circ } = 4$ $\left[ {\because p = OM = 4} \right]$
$x\left( {\dfrac{{\sqrt 3 }}{2}} \right) + y\left( {\dfrac{1}{2}} \right) = 4$
Simplifying it further,
$x\sqrt 3 + y + 8 = 0$
This is the required equation of a line.
Hence, the correct option is (A).
Note:The key concept to solve this type of question is to have a basic knowledge of properties (Angle sum property of triangle, Linear pair property) learned in previous classes. The figure should be made with proper attention and labeled correctly. The value of trigonometric angles should be known.
Recently Updated Pages
If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Find the cubic polynomial whose zeroes are 3 5 and class 11 maths JEE_Main

During the sale colour pencils were being sold in -class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In a class of 60 students 25 students play cricket class 11 maths JEE_Main

A regular polygon has 20 sides How many triangles can class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

Hybridisation in Chemistry – Concept, Types & Applications

