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(D) The intersection points of the given lines are the midpoints of the sides of triangle $ABC$. Let these points be $D(1, 2)$,$E(7, 5)$,and $F(2, 3)$

Vedclass · Accepted Answer

$6$

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(D) The intersection points of the given lines are the midpoints of the sides of triangle $ABC$. Let these points be $D(1, 2)$,$E(7, 5)$,and $F(2, 3)$.The area of the triangle formed by the midpoints $(\Delta DEF)$ is given by:$\Delta DEF = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$\Delta DEF = \frac{1}{2} |1(5 - 3) + 7(3 - 2) + 2(2 - 5)|$$\Delta DEF = \frac{1}{2} |1(2) + 7(1) + 2(-3)|$$\Delta DEF = \frac{1}{2} |2 + 7 - 6| = \frac{1}{2} |3| = 1.5$The area of the original triangle $ABC$ is $4$ times the area of the triangle formed by joining the midpoints:$	ext{Area}(ABC) = 4 	imes \Delta DEF = 4 	imes 1.5 = 6$.

Let the points of intersection of the lines $x-y+1=0$,$x-2y+3=0$,and $2x-5y+11=0$ be the midpoints of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ is .... .

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