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Question

Find the area of ABC with vertices A(0, −1), B(2, 1) and C(0, 3). Also, find the area of the triangle formed
by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1. [CBSE 2014]

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Solution

Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. Then
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y2 =1201-3+23+1+0-1-1 =12×8=4 sq. units
So, the area of the triangle ABC is 4 sq. units.
Let D(a1, b1), E(a2, b2) and F(a3, b3) be the midpoints of AB, BC and AC respectively. Then
a1=0+22=1 b1=-1+12=0a2=2+02=1 b2=1+32=2a3=0+02=0 b3=-1+32=1
Thus, the coordinates of D, E and F are D(a1 = 1, b1 = 0), E(a2 = 1, b2 = 2) and F(a3 = 0, b3 = 1). Now
AreaDEF=12a1b2-b3+a2b3-b1+a3b1-b2 =1212-1+11-0+00-2 =121+1+0=1 sq. unit
So, the area of the triangle DEF is 1 sq. unit.
Hence, ABC : DEF=4 : 1.

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