A(7, –3), B(5, 3) and C(3, –1) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides ΔABC into two triangles of equal areas.
D = {(3 + 5)/2,(3–1)/2} = (4,1)
For triangle ABD
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(7(3–1) + 5(1 + 3) + 4(–3–3))
= 1/2(14 + 20–24)
= 1/2(10)
= 5 sq unit
For triangle ACD
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(7(–1–1) + 3(1 + 3) + 4(–3 + 1))
= 1/2(–14 + 12–8)
= 1/2(10)
= 5 sq unit
Hence area of triangle ABD and ACD is equal.
7