Given: vertices of ΔABC = A( -5 , -1 ) , B( 3 , -5 ) & C( 5 , 2 )

To Prove: area of ΔABC = 4 times are of triangle formed by joining mid points of ΔABC

Let say the triangle formed by joining the vertices is ΔPQR

Formulas use are the following,

Area\:of\:Triangle=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|AreaofTriangle=

2

1 ∣x 1

(y 2

−y 3 )+x 2

(y 3 −y 1 )+x 3

(y 1 −y 2 )∣

Coordinates\:of\:Mid-Point=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})CoordinatesofMid−Point=(

2

x

1

+x

2

,

2

y

1

+y

2

)

Now,

Area\:of\:\Delta\,ABC=\frac{1}{2}\left|-5(-5-2)+3(2+1)+5(-1+5)\right|AreaofΔABC=

2

1

∣−5(−5−2)+3(2+1)+5(−1+5)∣

=\frac{1}{2}\left|-5(-7)+3(3)+5(4)\right|=

2

1

∣−5(−7)+3(3)+5(4)∣

=\frac{1}{2}\left|35+9+20\right|=

2

1

∣35+9+20∣

=\frac{1}{2}\times64=

2

1

×64

=32\:unit^2=32unit

2

Mid point of AB = P , Mid Point of CB = Q & Mid Point of AC = R

Coordinate of P = (\frac{-5+3}{2},\frac{-1-5}{2})(

2

−5+3

,

2

−1−5

)

= (\frac{-2}{2},\frac{-6}{2})(

2

−2

,

2

−6

)

= (-1,-3)(−1,−3)

Coordinate of Q = (\frac{5+3}{2},\frac{2-5}{2})(

2

5+3

,

2

2−5

)

= (\frac{8}{2},\frac{-3}{2})(

2

8

,

2

−3

)

= (4,\frac{-3}{2})(4,

2

−3

)

Coordinate of R = (\frac{-5+5}{2},\frac{-1+2}{2})(

2

−5+5

,

2

−1+2

)

= (\frac{0}{2},\frac{1}{2})(

2

0

,

2

1

)

= (0,\frac{1}{2})(0,

2

1

)

text\Area\:of\:\Delta\,PQR=\frac{1}{2}\left|-1(\frac{-3}{2}-\frac{1}{2})+4(\frac{1}{2}+3)+0(-3+\frac{-3}{2})\right|textAreaofΔPQR=

2

1

∣

∣

∣

−1(

2

−3

−

2

1

)+4(

2

1

+3)+0(−3+

2

−3

)

∣

∣

∣

=text\frac{1}{2}\left|-1(\frac{-4}{2})+4(\frac{1+6}{2})+0\right|text=

2

1

∣

∣

∣

−1(

2

−4

)+4(

2

1+6

)+0

∣

∣

∣

=text\frac{1}{2}\left|-1(-2)+2(7)+0\right|text=

2

1

∣−1(−2)+2(7)+0∣

=text\frac{1}{2}\left|2+14+0\right|text=

2

1

∣2+14+0∣

=8\:unit^2=8unit

2

Therefore, ar ΔABC = 4 × ΔPQR

Hence PROVED,