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Question
Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.
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Solution
We have, A (0, 4), B (1, 2) and C (3, 3)
Now, \[m_1 = \text { Slope of }AB = \frac{2 - 4}{1 - 0} = - 2\]
\[m_2 =\text { Slope of BC } = \frac{3 - 2}{3 - 1} = \frac{1}{2}\]
\[m_3 = \text { Slope of CA } = \frac{4 - 3}{0 - 3} = - \frac{1}{3}\]
\[\therefore m_1 m_2 = - 2 \times \frac{1}{2} = - 1\]
Therefore, AB is perpendicular to BC, i.e.
\[\angle ABC = {90}^\circ\].
Thus, the given points are the vertices of a right angled triangle.
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