Question

the vertices of a triangle ABC are A(- 5, 7) B(3,-5) C(5, 2). Show that the area of the triangle ABC is 4 times the area of the triangle formed by joining the midpoints of the sides of the triangle ABC​

Answer 1

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Answer 2

ANSWER:-

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Area of ΔABC whose vertices are is

$(x_1,y_1),(x_2,y_2) \: and \: (x_3,y_3)are$

$Area \: of ΔABC = \frac{1}{2} [x_1(y_2,y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

In ΔABC, vertices are A(−5,7),B(−4,−5) and C(4,5)

$Area \: of \: triangle \: = \frac{1}{2} (- 5 (- 5 - 5 ) - 4(5 - 7) + 4(7 + 5))$

$= \frac{1}{2} ( - 50 + 8 + 48)$

$= 5 \: sq. \: units$

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