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Question

A(7, −3), B(5, 3), C(3, −1) are the vertices of a ABC and AD is its median. Prove that the median
AD divides ABC into two triangles of equal areas.

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Solution

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).
Coordinates of D=5+32,3-12=4, 1
For the area of the triangle ADC, let Ax1,y1=A7,-3, Dx2,y2=D4,1 and Cx3,y3=C3,-1. Then
Area of ADC=12x1y2-y3+x2y3-y1+x3y1-y2 =1271+1+4-1+3+3-3-1 =1214+8-12=5 sq. unit
Now, for the area of triangle ABD, let Ax1, y1=A7,-3, Bx2, y2=B5, 3 and Dx3, y3=D4, 1. Then
Area of ABD=12x1y2-y3+x2y3-y1+x3y1-y2 =1273-1+51+3+4-3-3 =1214+20-24=5 sq. unit
Thus, AreaADC=AreaABD=5 sq. units.
Hence, AD divides ABC into two triangles of equal areas.

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