Question

# The points (a, a), (−a, a) and $\left(-\sqrt{3}a,\sqrt{3}a\right)$ form the vertices of (a) an equilateral triangle (b) a scalene triangle (c) an isosceles triangle (d) a right triangle

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Solution

## (a) an equilateral triangle Let A(a, a), B(−a, −a) and $C\left(-\sqrt{3}a,\sqrt{3}a\right)$ be the given points. Then, $AB=\sqrt{{\left(-a-a\right)}^{2}+{\left(-a-a\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-2a\right)}^{2}+{\left(-2a\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4{a}^{2}+4{a}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{8{a}^{2}}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}a\text{units}\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(-\sqrt{3}a+a\right)}^{2}+{\left(\sqrt{3}a+a\right)}^{2}}\phantom{\rule{0ex}{0ex}}=a\sqrt{{\left(1-\sqrt{3}\right)}^{2}+{\left(1+\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=a\sqrt{2\left(1+3\right)}\phantom{\rule{0ex}{0ex}}=a\sqrt{8}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}a\text{units}\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(-\sqrt{3}a-a\right)}^{2}+{\left(\sqrt{3}a-a\right)}^{2}}\phantom{\rule{0ex}{0ex}}=a\sqrt{{\left(\sqrt{3}+1\right)}^{2}+{\left(\sqrt{3}-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=a\sqrt{2\left(3+1\right)}\phantom{\rule{0ex}{0ex}}=a\sqrt{8}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}a\text{units}$ AB = BC = AC Hence, the points form the vertices of an equilateral triangle.

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