Question

# The points A(−4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is (a) isosceles (b) equilateral (c) scalene (d) right-angled

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Solution

## (a) isosceles Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then, $AB=\sqrt{{\left(4+4\right)}^{2}+{\left(0-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(8\right)}^{2}+{\left(0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{64+0}\phantom{\rule{0ex}{0ex}}=\sqrt{64}\phantom{\rule{0ex}{0ex}}=8\text{units}\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(0-4\right)}^{2}+{\left(3-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-4\right)}^{2}+{\left(3\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+9}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5\text{units}\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(0+4\right)}^{2}+{\left(3-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(4\right)}^{2}+{\left(3\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+9}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5\text{units}$ BC = AC = 5 units Therefore, $∆ABC$ is isosceles.

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