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# 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 (a) x − 3y + 1 = 0 (b) x − 3y + 4 = 0 (c) 3x − y + 2 = 0 (d) none of these

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Solution

## (b) x$-$3y = 4 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. $\therefore \mathrm{Slope}\mathrm{of}AC=-3$ $\mathrm{Thus},\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{altitude}\mathrm{through}B\mathrm{is}\frac{1}{3}.$ Equation of the required altitude is given below: $y-1=\frac{1}{3}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒x-3y+4=0$

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