In Fig. 6, ABC is a triangle coordinates of whose vertex A are (0, −1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF.
Solution
Let the coordinates of B and C be (x2, y2) and (x3, y3), respectively.
D is the midpoint of AB.
So,
`(1,0)=((x_2+0)/2,(y_2-1)/2)`
`=>1 = x_2/2 `
⇒ x2 = 2 and y2 = 1
Thus, the coordinates of B are (2, 1).
Similarly, E is the midpoint of AC.
So,
`(0,1)=((x_3+0)/2, (y_3-1)/2)`
`=>0=x_3/2 `
⇒ x3 = 0 and y3 = 3
Thus, the coordinates of C are (0, 3).
Also, F is the midpoint of BC. So, its coordinates are
`((2+0)/2,(1+3)/2)=(1,2)`
Now,
Area of a triangle = `1/2`[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
Thus, the area of ∆ABC is
`1/2[0(1-3)+2(3+1)+0(-1-1)]`
`=1/2xx8`
=4 square units
And the area of ∆DEF is
`1/2`[1(1−2)+0(2−0)+1(0−1)]
`=1/2xx(-2)`
=1 square unit (Taking the numerical value, as the area cannot be negative)
