An Important Question of class 10 Based on Equation of a Straight Line Chapter of M.L Aggarwal for ICSE BOARD.

Here vertices of a triangle are given.Find the equation of the altitude through a point.

This is the Question Number 32, Exercise 12.2 of M.L Aggarwal.

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# The vertices of a triangle are A (10, 4), B (4, – 9) and C (– 2, – 1). Find the equation of the altitude through A. The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.

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Given, vertices of a triangle are A (10, 4), B (4, – 9) and C (– 2, – 1)

Now,

Slope of line BC (m

_{1}) = (-1 + 9)/ (-2 – 4) = 8/ (-6) = -4/3Let the slope of the altitude from A (10, 4) to BC be m

_{2}Then, m

_{1}x m_{2}= -1(-4/3) x m

_{2}= -1m

_{2}= ¾So, the equation of the line will be

y – 4 = ¾ (x – 10)

4y – 16 = 3x – 30

3x – 4y – 14 = 0