Three sides AB, BC and CA of a triangle ABC are $5 x - 3 y + 2 = 0, x - 3 y - 2 = 0$ and $x + y - 6 = 0$ respectively. Find the equation of the altitude through the vertex A.

Q. The equation of perpendicular bisectors of the sides AB and AC of a triangle ABC are x - y + 5 = 0 and x + 2y = 0 respectively. If the point A is (1, -2), then the equation of line BC is

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If the middle points of the sides BC,CA and AB of the triangle ABC be (1,3),(5,7) and (−5,7), then the equation of the side AB is

The correct option is B x−y+12=0The line AB will be parallel to the line joining midpoints of AC and BC ∴ Slope of AB =y2−y1x2−x1=7−35−1=44=1The equation of AB:y−7=1(x+5)∴x−y+12=0

If the middle points of the sides $B C, C A$ and $A B$ of the triangle $A B C$ be $(1, 3), (5, 7)$ and $(- 5, 7)$ , then the equation of the side $A B$ is

The correct option is B $x - y + 12 = 0$
The line $A B$ will be parallel to the line joining midpoints of $A C$ and $B C$

$∴$ Slope of AB $= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{7 - 3}{5 - 1} = \frac{4}{4} = 1$

The equation of $A B$

$: y - 7 = 1 (x + 5)$

$∴ x - y + 12 = 0$