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Sia ? 6 years, 6 months ago
Let A (4, 4), B (3, 5) and C(-1, -1) be three vertices of a {tex}\Delta {\rm A}{\rm B}C{/tex}.
{tex}\therefore{/tex} Slope of AB {tex} = \frac{{5 - 4}}{{3 - 4}} = \frac{1}{{ - 1}} = - 1{/tex}
{tex}\therefore{/tex} Slope of BC {tex}= \frac{{ - 1 - 5}}{{ - 1 - 3}} = \frac{{ - 6}}{{ - 4}} = \frac{3}{2}{/tex}
{tex}\therefore{/tex} Slope of AC {tex} = \frac{{ - 1 - 4}}{{ - 1 - 4}} = \frac{{ - 5}}{{ - 5}} = 1{/tex}
Now slope of AB {tex}\times{/tex} slope of AC = -1{tex}\times{/tex}1 = -1
This shows thatAB{tex}\bot{/tex}AC. Thus {tex}\Delta {\rm A}{\rm B}C{/tex} is right angled at point A.
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