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Question 6
The vertices of triangle ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that ADAB=AEAC=14. Calculate the area of triangle ADE and compare it with area of triangle ABC.


Solution


Point D divides AB in ratio 1 : 3
Point E divides AC in the same ratio.
Hence, m1 = 1 and m2 = 3
Coordinates of D can be calculated as follows:
x=m1x2+m2x1m1+m2
=3×4+1×14=134
y=m1y2+m2y1m1+m2
=3×6+1×54=234
Coordinates of E can be calculated as follows:
x=1×7+3×44=194
y=1×2+3×64=5
Area of triangle ABC can be calculated as follows:
=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]
=12[4(52)+1(26)+7(65)]
=12(124+7)=152sq unit
Area of triangle ADE can be calculated as follows:
=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]
 =12[4(2345)+134(56)+194(6234)]
=12[4(23204)134+194(24234)]
=12(3134+1916)
=12[4852+1916]
=12×1516=1532sq unit
Hence, ratio of area of triangle ADE to area of triangle ABC
= 1532152 = 116  = 1 : 16

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