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Question
Find the coordinates of the orthocentre of the triangle whose vertices are (−1, 3), (2, −1) and (0, 0).
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Solution
Let A (0, 0), B (−1, 3) and C (2, −1) be the vertices of the triangle ABC.
Let AD and BE be the altitudes.
\[AD \perp BC\] and \[BE \perp AC\]
\[\therefore\] Slope of AD \[\times\] Slope of BC = −1
Slope of BE \[\times\] Slope of AC = −1
Here, slope of BC = \[\frac{- 1 - 3}{2 + 1} = - \frac{4}{3}\] and slope of AC = \[\frac{- 1 - 0}{2 - 0} = - \frac{1}{2}\]
\[\therefore \text { Slope of AD } \times \left( - \frac{4}{3} \right) = - \text { 1 and slope of BE } \times \left( - \frac{1}{2} \right) = - 1 \]
\[ \Rightarrow \text { Slope of AD } = \frac{3}{4}\text { and slope of BE } = 2\]
The equation of the altitude AD passing through A (0, 0) and having slope \[\frac{3}{4}\] is
\[y - 0 = \frac{3}{4}\left( x - 0 \right)\]
\[ \Rightarrow y = \frac{3}{4}x . . . . (1)\]
The equation of the altitude BE passing through B (−1, 3) and having slope 2 is
\[y - 3 = 2\left( x + 1 \right)\]
\[ \Rightarrow 2x - y + 5 = 0 . . . . (2)\]
Solving (1) and (2):
x = − 4, y = − 3
Hence, the coordinates of the orthocentre is (−4, −3).
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