(C) Let the coordinates of point $P$ be $(x, y)$. The area of $\triangle PAB$ is given by the determinant formula: $\frac{1}{2} |x(3-5) + 2(5-y) + (-4

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(C) Let the coordinates of point $P$ be $(x, y)$. The area of $\triangle PAB$ is given by the determinant formula: $\frac{1}{2} |x(3-5) + 2(5-y) + (-4)(y-3)| = 12$ $\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$ $|-2x - 6y + 22| = 24$ $|x + 3y - 11| = 12$ This implies $x + 3y - 11 = 12$ or $x + 3y - 11 = -12$. Thus,the locus is $(x + 3y - 23)(x + 3y + 1) = 0$. Expanding this: $x^2 + 3xy + x + 3xy + 9y^2 + 3y - 23x - 69y - 23 = 0$ $x^2 + 6xy + 9y^2 - 22x - 66y - 23 = 0$.

Class 11 Mathematics Straight Lines Locus of Point

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If $A=(2,3)$ and $B=(-4,5)$ are two fixed points,then the locus of a point $P$ such that the area of $\triangle PAB$ is $12$ square units is

AP EAMCET 2023 All AP EAMCET

A
$x^2+6xy+9y^2+22x+66y+23=0$
B
$x^2-6xy+9y^2+22x+66y+23=0$
C
$x^2+6xy+9y^2-22x-66y-23=0$
D
$x^2-6xy+9y^2-22x-66y-23=0$

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If $A=(2,3)$ and $B=(-4,5)$ are two fixed points,then the locus of a point $P$ such that the area of $\triangle PAB$ is $12$ square units is

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