Let the mirror image of point A(1, 4) in the | vedclass

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(A) $1$. The mirror image of point $(x, y)$ in the line $y = x$ is $(y, x)$. Thus,for $A(1, 4)$,$B = (4, 1)$.$2$. The mirror image of point $(x, y)$ i

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$3$

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(A) $1$. The mirror image of point $(x, y)$ in the line $y = x$ is $(y, x)$. Thus,for $A(1, 4)$,$B = (4, 1)$.$2$. The mirror image of point $(x, y)$ in the line $y = -x$ is $(-y, -x)$. Thus,for $B(4, 1)$,$C = (-1, -4)$.$3$. The mirror image of point $(x, y)$ in the $x$-axis is $(x, -y)$. Thus,for $C(-1, -4)$,$D = (-1, 4)$.$4$. The coordinates are $A(1, 4)$,$B(4, 1)$,and $D(-1, 4)$.$5$. The area of $	riangle ABD$ with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.$6$. Area $= \frac{1}{2} |1(1 - 4) + 4(4 - 4) + (-1)(4 - 1)| = \frac{1}{2} |1(-3) + 4(0) - 1(3)| = \frac{1}{2} |-3 - 3| = \frac{1}{2} |-6| = 3 \, sq. \, units$.

Let the mirror image of point $A(1, 4)$ in the line $y = x$ be point $B$; the mirror image of point $B$ in the line $y = -x$ be $C$; and the mirror image of $C$ in the $x$-axis be $D$. Then,the area of triangle $ABD$ is ............... $sq. \, units$.

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