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First to find the area we need to find the vertices of the triangle.

Midpoint D = x1+x2 = -1........(1)

y1+y2 = 5.............................(4)

Midpoint E = x1+x3 = 14.........(2)

y1+y3 = 6...............................(5)

Midpoint F = x2+x3 = 7............(3)

y2+y3 = 7...............................(6)

Now let us find the x terms first.

(1) ........ x1 = -1-x2 Sub in (2) ........ -1-x2 + x3 = 14........(7) Now subtracting (7) and (3) we get, 2x3 = 22 x3 = 11 Sub x3 in (2) ........x1+11 = 14 x1 = 3 and so x2 = -4

now let us find y terms.

(3) ............y1 = 5 - y2 sub in (5).........5-y2+y3 = 6.........(8) Now subtracting (8) and (6) we get, y3 = 4 now sub y3 in (5)...... y1 = 2 and so y2 = 3

now the vertices are (3,2) (-4,3) (11,4)

Area of the triangle = (3(3-4) + (-4) (4-2) + 11(2-3))/2 = (-3 -8 -11)2 = - 11

Remove the negative sign (-) from the number -11.

The area of the triangle is 11 sq.units.

Because the formula calls for absolute value, you simply remove the negative sign.

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If DEF is the triangle formed by joining the mid-points of sides of the triangle ABC, then the area of triangle ABC is 4 times the area of triangle DEF. So find the area of triangle DEF first and multiply with 4 to get the area of ABC.

Area of a triangle with vertices (x₁,y₁); (x₂,y₂) and (x₃,y₃) is given by Area = [ x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) ] / 2