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 \[x=\frac{1-\sqrt{3}}{2},\,\,\,\,\,y=\frac{5\sqrt{3}+7}{2}\]You need to login to perform this action.
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