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Question
The equations of the sides AB, BC and CA of ∆ ABC are y − x = 2, x + 2y = 1 and 3x + y + 5 = 0 respectively. The equation of the altitude through B is
Options
x − 3y + 1 = 0
x − 3y + 4 = 0
3x − y + 2 = 0
none of these
Solution
x − 3y + 4 = 0
The equation of the sides AB, BC and CA of ∆ABC are y − x = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively.
Solving the equations of AB and BC, i.e. y − x = 2 and x + 2y = 1, we get:
x = − 1, y = 1
So, the coordinates of B are (−1, 1).
The altitude through B is perpendicular to AC.
\[\therefore \text { Slope of AC} = - 3\]
\[\text { Thus, slope of the altitude through B is } \frac{1}{3} .\]
Equation of the required altitude is given below:
\[y - 1 = \frac{1}{3}\left( x + 1 \right)\]
\[ \Rightarrow x - 3y + 4 = 0\]
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