prove that the area of a triangle whose vertices are (-5,1) , (3,2) AND (5,7) is 4 times the word area of the triangle formed by the lines joining midpoints of the sides
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Correct option is
C
32 square units
Area of a triangle with vertices (x
1
,y
1
) ; (x
2
,y
2
) and (x
3
,y
3
) is
∣
∣
∣
∣
∣
2
x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)
∣
∣
∣
∣
∣
Hence, substituting the points (x
1
,y
1
)=(−5,−1) ; (x
2
,y
2
)=(3,−5) and (x
3
,y
3
)=(5,2) in the area formula, we get
Area of triangle ABC =
∣
∣
∣
∣
∣
2
(−5)(−5−2)+(3)(2+1)+5(−1+5)
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
2
35+9+20
∣
∣
∣
∣
∣
=
2
64
=32squnits
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