Coordinate Geometry

Question

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A(4, - 6), B(3,- 2) and C(5, 2) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides ΔABC into two triangles of equal areas.

Solution

Let co-ordinate of D(x,y) and D is mid point of BC

$x = \frac{(3 + 5)}{2} = 4 y = \frac{(2 - 2)}{2} = 0$

$Now area of triangle ABD = \frac{1}{2} {4 (- 2 - 0) + 3 [(0 - (- 6))] + 4 [(- 6) - (- 2)]} = 0.5 x [- 8 + 18 - 16] = 3 sq . unit And area of triangle ACD = \frac{1}{2} [5 (- 6 - 0) + 4 (0 - 2) + 4 (2 + 6)] = 0.5 x [- 30 + (- 8) + 32] = 3 sq . unit$

Hence, the median AD divides triangle ABC into two triangles of equal area.

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Coordinate Geometry

A(4, - 6), B(3,- 2) and C(5, 2) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides ΔABC into two triangles of equal areas.

Some More Questions From Coordinate Geometry Chapter

Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6), are equal and bisect each other.

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