question_answer

The coordinates of the third vertex of an equilateral triangle whose two vertices are at \[(3,\,4),\,\,(-2,\,3)\] are _____.

A)  \[(1,7)\]        

B)  \[(5,1)\]

C)  \[\left( \frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2} \right)\] or \[\left( \frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2} \right)\]

D)  \[(-5,5)\]

Correct Answer: C

Solution :

Since \[\Delta ABC\] is equilateral triangle \[=\frac{\sqrt{3}}{4}\times {{(side)}^{2}}\] \[\therefore \]      \[AB=AC\Rightarrow A{{B}^{2}}=A{{C}^{2}}\] \[\Rightarrow \]      \[{{(x-3)}^{2}}+{{(y-4)}^{2}}\]             \[={{(x+2)}^{2}}+{{(y-3)}^{2}}\] \[\Rightarrow \]            \[5x+y-6=0\]              ?..(i) Day Now, area of equilateral triangle \[=\frac{\sqrt{3}}{4}\times {{(side)}^{2}}\] \[\therefore \]  Area of \[\Delta ABC=\frac{\sqrt{3}}{4}\times {{(BC)}^{2}}=\frac{13\sqrt{3}}{2}\]units. \[\Rightarrow \] \[\frac{13\sqrt{3}}{2}=\frac{1}{2}|x(1)+3(3-y)-2(y-4)|\] \[\Rightarrow \]            \[\pm 13\sqrt{3}=x+9-3y-2y+8\] \[\Rightarrow \]            \[x-5y+17=\pm 13\sqrt{3}\]               \[\Rightarrow \]            \[x-5y=13\sqrt{3}-17\]               ??(ii) \[\Rightarrow \]            \[x-5y=-\left( 13\sqrt{3}+17 \right)\]     ?..(iii) Solving (i) and (ii), we get \[x=\frac{1-\sqrt{3}}{2},\,y=\frac{7-5\sqrt{3}}{2}\] Also, on solving (i) and (ii), we get \[x=\frac{1-\sqrt{3}}{2},\,\,\,\,\,y=\frac{5\sqrt{3}+7}{2}\]