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

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(C) Let $P = (x, y)$.Given $A = (2, 3)$ and $B = (-4, 5)$.The area of $\Delta PAB = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 1

Vedclass · Accepted Answer

$x^2 + 6xy + 9y^2 - 22x - 66y - 23 = 0$

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(C) Let $P = (x, y)$.Given $A = (2, 3)$ and $B = (-4, 5)$.The area of $\Delta PAB = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12$.Substituting the coordinates: $\frac{1}{2} |x(3 - 5) + 2(5 - y) - 4(y - 3)| = 12$.$\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$.$|-2x - 6y + 22| = 24$.$|-x - 3y + 11| = 12$.This implies $-x - 3y + 11 = 12$ or $-x - 3y + 11 = -12$.Case $1$: $x + 3y + 1 = 0$.Case $2$: $x + 3y - 23 = 0$.The combined equation of the locus is $(x + 3y + 1)(x + 3y - 23) = 0$.Expanding this: $x^2 + 3xy - 23x + 3xy + 9y^2 - 69y + x + 3y - 23 = 0$.$x^2 + 6xy + 9y^2 - 22x - 66y - 23 = 0$.

Let $A (2, 3)$ and $B (-4, 5)$ be two fixed points. $A$ point $P$ moves in such a way that the area of $\Delta PAB = 12 \, \text{sq. units}$. Find its locus.

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