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Question
Find the orthocentre of the triangle the equations of whose sides are x + y = 1, 2x + 3y = 6 and 4x − y + 4 = 0.
Solution
The given lines are as follows:
x + y = 1 ... (1)
2x + 3y = 6 ... (2)
4x − y + 4 = 0 ... (3)
In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.
Solving (1) and (2):
x = −3, y = 4
Thus, AB and BC intersect at B (−3, 4).
Solving (1) and (3):
x = \[- \frac{3}{5}\] , y = \[\frac{8}{5}\]
Thus, AB and CA intersect at
\[A \left( - \frac{3}{5}, \frac{8}{5} \right)\].
Let AD and BE be the altitudes.
\[AD \perp BC \text { and }BE \perp AC\]
\[\therefore\] Slope of AD \[\times\] Slope of BC = −1
and Slope of BE \[\times\] Slope of AC = −1
Here, slope of BC = slope of the line (2) = \[- \frac{2}{3}\] and slope of AC = slope of the line (3) = 4
\[\therefore \text { Slope of AD } \times \left( - \frac{2}{3} \right) = - 1 \text { and slope of BE } \times 4 = - 1\]
\[ \Rightarrow \text { Slope of AD } = \frac{3}{2} \text { and slope of BE } = - \frac{1}{4}\]
The equation of the altitude AD passing through \[A \left( - \frac{3}{5}, \frac{8}{5} \right)\] and having slope \[\frac{3}{2}\] is \[y - \frac{8}{5} = \frac{3}{2}\left( x + \frac{3}{5} \right)\]
\[\Rightarrow 3x - 2y + 5 = 0\] ... (4)
The equation of the altitude BE passing through B (−3, 4) and having slope \[- \frac{1}{4}\] is \[y - 4 = - \frac{1}{4}\left( x + 3 \right)\]
\[\Rightarrow x + 4y - 13 = 0\] ... (5)
Solving (4) and (5), we get
\[\left( \frac{3}{7}, \frac{22}{7} \right)\] as the orthocentre of the triangle.
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