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Question

If D, E and F are the mid-points of sides BC, CA and AB respectively of a ΔABC, then using coordinate geometry prove that Area of ΔDEF=14(AreaofΔABC)


Solution

Concept:
Application:

Let A(x1,y1),B(x2,y2),C(x3,y3) be the vertices of  ΔABC. Then, the coordinates of D, E and F are x2+x32,(y2+y32),x1+x32,y1+y32) and (x1+x22,y1+y22) respectively.

Δ1=Area of ΔABC=12|x1(y2y3)+x2(y3y1)+x3(y1y2)|

Δ2=Area of ΔDEF=12(x2+x32)(y1+y32y1+y22)+(x1+x32)(y1+y22y2+y32)+(x1+x22)(y2+y32y1+y32)

Δ2=18|(x2+x3)(y3y2)+(x1+x3)(y1y3)+(x1+x2)(y2y1)|

Δ2=18|x1(y1)(y3+y2y1)+x2(y3y2+y2+y1)+x3(y3y2+y1y3)|

Δ2=18|x1(y2y3)+x2(y3y1)+x3(y1y2)|

Δ2=14 (Area of ΔABC)=14Δ1

Hence, Area of ΔDEF=14      (Area of ΔABC)

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