Find the area of the triangle ABC with vertex | vedclass

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(D) Let the coordinates of vertices $B$ and $C$ be $(a, b)$ and $(x, y)$ respectively.The mid-point of side $AB$ is given as $(2, -1)$. Thus,$\left(\f

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$12$

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(D) Let the coordinates of vertices $B$ and $C$ be $(a, b)$ and $(x, y)$ respectively.The mid-point of side $AB$ is given as $(2, -1)$. Thus,$\left(\frac{1+a}{2}, \frac{-4+b}{2}ight) = (2, -1)$.Equating the coordinates,we get $1+a = 4 \implies a = 3$ and $-4+b = -2 \implies b = 2$. So,$B = (3, 2)$.The mid-point of side $AC$ is given as $(0, -1)$. Thus,$\left(\frac{1+x}{2}, \frac{-4+y}{2}ight) = (0, -1)$.Equating the coordinates,we get $1+x = 0 \implies x = -1$ and $-4+y = -2 \implies y = 2$. So,$C = (-1, 2)$.The vertices of $	riangle ABC$ are $A(1, -4)$,$B(3, 2)$,and $C(-1, 2)$.The area of $	riangle ABC = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |1(2 - 2) + 3(2 - (-4)) + (-1)(-4 - 2)|$.Area $= \frac{1}{2} |0 + 3(6) - 1(-6)| = \frac{1}{2} |18 + 6| = \frac{24}{2} = 12$ $sq. units$.

Find the area of the triangle $ABC$ with vertex $A (1, -4)$ and the mid-points of the sides passing through $A$ being $(2, -1)$ and $(0, -1)$ (in $sq. units$).

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