Let A(2, 3) and B(-4, 5) be two fixed points. | vedclass

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(A) The area of a triangle 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_

Vedclass · Accepted Answer

$x + 3y + 1 = 0$ or $x + 3y - 23 = 0$

Vedclass · Answer

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12$. Substituting $A(2, 3), B(-4, 5)$,and $P(x, y)$: $\frac{1}{2} |x(3 - 5) + 2(5 - y) + (-4)(y - 3)| = 12$ $\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$ $|-2x - 6y + 22| = 24$ $|-x - 3y + 11| = 12$ This gives two cases: Case $1: -x - 3y + 11 = 12 \implies x + 3y + 1 = 0$ Case $2: -x - 3y + 11 = -12 \implies x + 3y - 23 = 0$ Thus,the locus is $x + 3y + 1 = 0$ or $x + 3y - 23 = 0$.

Let $A(2, 3)$ and $B(-4, 5)$ be two fixed points. If a point $P(x, y)$ moves such that the area of $\Delta PAB = 12$ square units,find the locus of $P$.

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