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 (m1) = (-1 + 9)/ (-2 – 4) = 8/ (-6) = -4/3
Let the slope of the altitude from A (10, 4) to BC be m2
Then, m1 x m2 = -1
(-4/3) x m2 = -1
m2 = ¾
So, the equation of the line will be
y – 4 = ¾ (x – 10)
4y – 16 = 3x – 30
3x – 4y – 14 = 0