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If (0, −3) and (0, 3) Are the Two Vertices of an Equilateral Triangle, Find the Coordinates of Its Third Vertex.

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Question

If (0, −3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.    

Answer in Brief
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Solution

Let the given points be A(0, −3) and B(0, 3). Suppose the coordinates of the third vertex be C(xy).

Now, ∆ABC is an equilateral triangle.

∴ AB = BC = CA

\[\sqrt{\left( 0 - 0 \right)^2 + \left( - 3 - 3 \right)^2} = \sqrt{\left( x - 0 \right)^2 + \left( y - 3 \right)^2} = \sqrt{\left( x - 0 \right)^2 + \left[ y - \left( - 3 \right) \right]^2}\]                  (Distance formula)

Squaring on both sides, we get   \[36 = x^2 + \left( y - 3 \right)^2 = x^2 + \left( y + 3 \right)^2\]
⇒ \[x^2 + \left( y - 3 \right)^2 = x^2 + \left( y + 3 \right)^2\]
\[x^2 + \left( y - 3 \right)^2 = 36\]
Now,

\[x^2 + \left( y - 3 \right)^2 = x^2 + \left( y + 3 \right)^2 \]

\[ \Rightarrow y^2 - 6y + 9 = y^2 + 6y + 9\]

\[ \Rightarrow - 12y = 0\]

\[ \Rightarrow y = 0\]

Putting y = 0 in

\[x^2 + \left( y - 3 \right)^2 = 36\] , we get

\[x^2 + \left( 0 - 3 \right)^2 = 36\]

\[ \Rightarrow x^2 = 36 - 9 = 27\]

\[ \Rightarrow x = \pm \sqrt{27} = \pm 3\sqrt{3}\]

Thus, the coordinates of the third vertex are

\[\left( 3\sqrt{3}, 0 \right)\]   or  \[\left( - 3\sqrt{3}, 0 \right)\]
 
 
 
 

 

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Chapter 6: Co-ordinate Geometry - Exercise 6.2 [Page 17]

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R.D. Sharma Mathematics [English] Class 10
Chapter 6 Co-ordinate Geometry
Exercise 6.2 | Q 48 | Page 17

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