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  • The coordinates of the end points of base  of an equilateral triangle <img alt="\mathrm{ABC}" s

The coordinates of the end points of base \mathrm{BC} of an equilateral triangle \mathrm{ABC} are \mathrm{B\left ( 1,3 \right )} and \mathrm{C\left ( -2,7 \right )} respectively. Find the coordinates of the vertex \mathrm{A}.

 

Option: 1

\mathrm{\left(-\frac{1}{2} \pm 2 \sqrt{3}, 5 \pm \frac{3 \sqrt{3}}{2}\right)}


Option: 2

\mathrm{\left(\frac{-1}{2} \pm \frac{3 \sqrt{3}}{2}, 5 \pm 2 \sqrt{3}\right)}


Option: 3

\mathrm{\left( \pm 2 \sqrt{3}, \pm 3 \frac{\sqrt{3}}{2}\right)}


Option: 4

\mathrm{\left(-\frac{1}{2} \pm \frac{2 \sqrt{3}}{3}, 5 \pm \frac{2}{\sqrt{2}}\right)}


Answers (1)

best_answer

We have, \mathrm{\mathrm{BC}=\sqrt{9+16}=5}
Let \mathrm{D} be the mid-point of base \mathrm{BC}. The coordinates of \mathrm{D} are \mathrm{\left ( -1/2,5 \right )}. The length \mathrm{AD} of the altitude through the vertex \mathrm{A} is given by \mathrm{\mathrm{AD}=\frac{\sqrt{3}(\mathrm{BC})}{2}=\frac{5 \sqrt{3}}{2}}|
Clearly, the vertex \mathrm{A} lies on a line passing through \mathrm{D} and perpendicular to \mathrm{BC} at a distance of \mathrm{5\sqrt{3}/2} from \mathrm{D}. We have Slope of \mathrm{\mathrm{BC}=(7-3) /(-2-1)=-4 / 3}
\mathrm{\therefore } Slope of \mathrm{AD = 3/4 }
suppose AD makes an angle θ with BC. Then, the equation of AD is \mathrm{\frac{x+1 / 2}{\cos \theta}=\frac{y-5}{\sin \theta} }
Thus, the coordinates of \mathrm{A } are \mathrm{(-1 / 2 \pm 2 \sqrt{3}, 5 \pm 3 \sqrt{3} / 2) }

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Anam Khan

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