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.

Answer 1

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

\[y - 4 = - 3\left( x - 2 \right)\]

\[ \Rightarrow 3x + y = 10\]