
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
219.9k+ 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.
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