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Question
Three sides AB, BC and CA of a triangle ABC are 5x − 3y + 2 = 0, x − 3y − 2 = 0 and x + y − 6 = 0 respectively. Find the equation of the altitude through the vertex A.
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Solution
The sides AB, BC and CA of a triangle ABC are as follows:
5x − 3y + 2 = 0 ... (1)
x − 3y − 2 = 0 ... (2)
x + y − 6 = 0 ... (3)
Solving (1) and (3):
x = 2 , y = 4
Thus, AB and CA intersect at A (2, 4).
Let AD be the altitude.
\[AD \perp BC\]
\[\therefore\] Slope of AD \[\times\] Slope of BC = −1
Here, slope of BC = slope of the line (2) = \[\frac{1}{3}\]
\[\therefore \text { Slope of AD }\times \frac{1}{3} = - 1 \]
\[ \Rightarrow \text { Slope of AD } = - 3\]
Hence, the equation of the altitude AD passing through A (2, 4) and having slope −3 is
\[y - 4 = - 3\left( x - 2 \right)\]
\[ \Rightarrow 3x + y = 10\]
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