Question

If (2,4), (2,6) are two vertices of an equilateral triangle then the third vertex is
(a) \[(2+\sqrt{3},\,5)\]
(b) \[(\sqrt{3}-2,\,5)\]
(c) \[(5,\,2+\sqrt{3})\]
(d) \[(5,\,2-\sqrt{3})\]

Answer 1

Hint: Draw the equilateral triangle with the given vertices and let the third vertex be (x, y). As it is an equilateral triangle all sides are of equal length and the angle between all sides is \[{{60}^{\circ }}\]. We will be using distance formula which is \[d=\sqrt{\left[ {{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}} \right]}\] where \[({{x}_{1}},\,{{y}_{1}})\] and \[({{x}_{2}},\,{{y}_{2}})\] are the coordinates of any two points.

Complete step-by-step answer:


Before proceeding with the question, we should know about equilateral triangles. An equilateral triangle is a triangle whose all sides are equal and the angle between all sides is \[{{60}^{\circ }}\].
Let the coordinates of A be (2, 4) and B be (2, 6) and C be (x, y).

We will first calculate the distance AB between coordinates A and B, then the distance BC between B and C and finally the distance AC between A and C
 \[AB=\sqrt{\left[ {{(2-2)}^{2}}+{{(6-4)}^{2}} \right]}=\sqrt{{{2}^{2}}}=2\,units.......(1)\]
 \[BC=\sqrt{\left[ {{(2-x)}^{2}}+{{(6-y)}^{2}} \right]}=2\,units........(2)\]
 \[AC=\sqrt{\left[ {{(2-x)}^{2}}+{{(4-y)}^{2}} \right]}=2\,units......(3)\]
We know that all the sides are equal so now \[AB=BC=AC\] .
Equating equation (2) and equation (3) that is \[BC=AC\] we get,
 \[\,\Rightarrow \sqrt{\left[ {{(2-x)}^{2}}+{{(6-y)}^{2}} \right]}=\sqrt{\left[ {{(2-x)}^{2}}+{{(4-y)}^{2}} \right]}...........(4)\]
Now squaring both sides in equation (4) we get,
 \[\,\Rightarrow {{(2-x)}^{2}}+{{(6-y)}^{2}}={{(2-x)}^{2}}+{{(4-y)}^{2}}.........(5)\]
Cancelling similar terms on both sides from equation (5) we get,
 \[\,\Rightarrow {{(6-y)}^{2}}={{(4-y)}^{2}}.........(6)\]
Now expanding the remaining terms in equation (6) we get,
 \[\,\Rightarrow 36+{{y}^{2}}-12y=16+{{y}^{2}}-8y........(7)\]
Cancelling similar terms on both sides from equation (7) and solving we get,
 \[\begin{align}
  & \,\Rightarrow 12y-8y=36-16 \\
 & \,\Rightarrow 4y=20 \\
 & \,\Rightarrow y=5......(8) \\
\end{align}\]
Now putting value of y from equation (8) in equation (2) we get,
 \[\,\Rightarrow \sqrt{\left[ {{(2-x)}^{2}}+{{(6-5)}^{2}} \right]}=2\,.......(9)\]
Squaring both sides and solving we get,
 \[\begin{align}
  & \,\Rightarrow {{(2-x)}^{2}}+{{(6-5)}^{2}}=4\, \\
 & \,\Rightarrow {{(2-x)}^{2}}+1=4 \\
 & \,\Rightarrow {{(2-x)}^{2}}=3......(10) \\
\end{align}\]
Taking square root of both sides of equation (10) and solving for x we get,
 \[\begin{align}
  & \,\Rightarrow (2-x)=\pm \sqrt{3} \\
 & \,\Rightarrow x=2\pm \sqrt{3} \\
\end{align}\]
Hence the possible coordinates of third vertex C are \[(2+\sqrt{3},\,5)\] and \[(2-\sqrt{3},\,5)\].
Hence the correct answer is option (a).

Note: We have to be very careful while using distance formula because in a hurry we may make simple calculation mistakes. Alternatively, we would have substituted the value of y in equation (3) then also we would have got the same answer.