Question

If one vertex of an equilateral triangle of side ‘a’ lies at the origin and the other lies on the line $ x - \sqrt 3 y = 0, $ then the coordinates of the third vertex are
This question has multiple correct options
A. $ (0,a) $
B. $ \left( {\dfrac{{\sqrt 3 a}}{2},\dfrac{a}{2}} \right) $
C. $ (0, - a) $
D. $ \left( {\dfrac{{ - \sqrt 3 a}}{2},\dfrac{a}{2}} \right) $

Answer 1

Hint: Here we will use the standard formula for the line and will find the slope of the given line and then accordingly will find the coordinate for the third vertex. Also, will look for the possibilities where the third vertex of the triangle lies.

Complete step-by-step answer:
Take the given line equation: $ x - \sqrt 3 y = 0 \Rightarrow x = \sqrt 3 y $
Make in the standard form-
 $ \Rightarrow \dfrac{x}{{\sqrt 3 }} = y $
Compare the given equation with the standard equation-
 $ y = mx + c $
Slope of the line: $ m = \dfrac{1}{{\sqrt 3 }} $
which is $ 30^\circ $ with positive x-axis.
Now, as the triangle is an equilateral the other vertex must be on \[x = - \sqrt 3 y\] .... (A)
Let us assume that the side be $ (x,y) $
Hence, $ {x^2} + {y^2} = {a^2} $
Place the values in the above equation from the equation (A)
 $ \Rightarrow {\left( {\sqrt 3 y} \right)^2} + {y^2} = {a^2} $
Simplify the above equation –
\[
   \Rightarrow 3{y^2} + {y^2} = {a^2} \\
   \Rightarrow 4{y^2} = {a^2} \\
 \]
Make required “y” the subject –
 $ \Rightarrow {y^2} = \dfrac{{{a^2}}}{4} $
Take square-root on both the sides of the equation –
 $ \Rightarrow \sqrt {{y^2}} = \sqrt {\dfrac{{{a^2}}}{4}} $
Square and square-root cancel each other on the left hand side of the equation – $ \Rightarrow y = \pm \dfrac{a}{2} $
Now, place values in the equation (A)
 $ \Rightarrow x = \pm \dfrac{{\sqrt 3 a}}{2} $
Therefore, if the triangle lies in the first and fourth quadrant then the vertex is $ \left( {\dfrac{{\sqrt 3 a}}{2},\dfrac{a}{2}} \right) $
And similarly if the triangle lies in the second and the third quadrant then the vertex is $ \left( {\dfrac{{ - \sqrt 3 a}}{2},\dfrac{{ - a}}{2}} \right) $
So, the correct answer is “Option B and D”.

Note: Remember the intersection of x-axis and the y-axis divides the coordinate plane into four sections and these four sections are known as quadrants. In quadrant I, both the x and y coordinates are positive, in quadrant II, both the x is negative and y coordinate is positive, in quadrant III, both the x and y coordinates are negative and in quadrant IV, both the x is positive and y is negative.