The locus of P such that the area of Delta PA | vedclass

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(B) Let $P$ be $(x, y)$. The area of $\Delta PAB$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12$.Substituting the c

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$(x + 3y + 1)(x + 3y - 23) = 0$

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(B) Let $P$ be $(x, y)$. The area of $\Delta PAB$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12$.Substituting the coordinates $P(x, y)$,$A(2, 3)$,and $B(-4, 5)$:$\frac{1}{2} |x(3 - 5) + 2(5 - y) + (-4)(y - 3)| = 12$$\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$$|-2x - 6y + 22| = 24$$|-x - 3y + 11| = 12$This implies $-x - 3y + 11 = 12$ or $-x - 3y + 11 = -12$.Case $1$: $x + 3y + 1 = 0$.Case $2$: $x + 3y - 23 = 0$.Therefore,the locus is $(x + 3y + 1)(x + 3y - 23) = 0$.

The locus of $P$ such that the area of $\Delta PAB = 12 \text{ sq. units}$,where $A(2, 3)$ and $B(-4, 5)$ is:

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