Let ABC be a triangle with A(–1, –5), B(0, 0) and C(2, 2) and let D be the middle point of BC. Find the equation of the perpendicular drawn from Bto AD.
x + 3y = 0
âˆµ D is the middle point of BC.
i.e., D(1, 1)
Slope of the median AD
∴ Slope of BM which is perpendicular to AD = –1/3.
Hence equation of the line BM is
x + 3y = 0
which is the required equation of the line.
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