B(2,3),C(5,-2),and D(1,-1) are three points. | vedclass

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(D) The area of a quadrilateral with vertices $A(x, y)$,$B(2, 3)$,$C(5, -2)$,and $D(1, -1)$ is given by the formula: $	ext{Area} = \frac{1}{2} |(x_A

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$(4x-y-42)(4x-y-2)=0$

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(D) The area of a quadrilateral with vertices $A(x, y)$,$B(2, 3)$,$C(5, -2)$,and $D(1, -1)$ is given by the formula: $	ext{Area} = \frac{1}{2} |(x_A y_B - y_A x_B) + (x_B y_C - y_B x_C) + (x_C y_D - y_C x_D) + (x_D y_A - y_D x_A)|$. Substituting the coordinates: $	ext{Area} = \frac{1}{2} |(3x - 2y) + (-4 - 15) + (-5 + 2) + (y - x)| = 10$. $\frac{1}{2} |2x - y - 22| = 10$. $|2x - y - 22| = 20$. This gives two lines: $2x - y - 22 = 20 \implies 2x - y - 42 = 0$ and $2x - y - 22 = -20 \implies 2x - y - 2 = 0$. However,checking the order of vertices $A, B, C, D$,the area is $\frac{1}{2} |(x(3) - y(2)) + (2(-2) - 3(5)) + (5(-1) - (-2)(1)) + (1(y) - (-1)(x))| = 10$. $\frac{1}{2} |3x - 2y - 4 - 15 - 5 + 2 + y + x| = 10$. $\frac{1}{2} |4x - y - 22| = 10$. $|4x - y - 22| = 20$. This results in $4x - y - 42 = 0$ or $4x - y - 2 = 0$. Thus,the locus is $(4x - y - 42)(4x - y - 2) = 0$.

$B(2,3)$,$C(5,-2)$,and $D(1,-1)$ are three points. If $A(x, y)$ is a variable point such that the area of the quadrilateral $ABCD$ is $10 \text{ sq. units}$,then the locus of $A$ is

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