Two points A and B with coordinates (1, 1) an | vedclass

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(A) Let the point $P$ be $(x, y)$. The coordinates of $A$ are $(1, 1)$ and $B$ are $(-2, 3)$.Given that the area of $	riangle PAB = 9 	ext{ sq. unit

Vedclass · Accepted Answer

$2x + 3y + 13 = 0 	ext{ and } 2x + 3y - 23 = 0$

Vedclass · Answer

(A) Let the point $P$ be $(x, y)$. The coordinates of $A$ are $(1, 1)$ and $B$ are $(-2, 3)$.Given that the area of $	riangle PAB = 9 	ext{ sq. units}$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the coordinates: $\frac{1}{2} |x(1 - 3) + 1(3 - y) + (-2)(y - 1)| = 9$.$\frac{1}{2} |-2x + 3 - y - 2y + 2| = 9$.$|-2x - 3y + 5| = 18$.This implies $-2x - 3y + 5 = 18$ or $-2x - 3y + 5 = -18$.Case $1$: $-2x - 3y = 13 \Rightarrow 2x + 3y + 13 = 0$.Case $2$: $-2x - 3y = -23 \Rightarrow 2x + 3y - 23 = 0$.Thus,the locus is $2x + 3y + 13 = 0$ and $2x + 3y - 23 = 0$.

Two points $A$ and $B$ with coordinates $(1, 1)$ and $(-2, 3)$ are given respectively. Then,the locus of a point $P$ such that the area of $\triangle PAB$ is $9 \text{ sq. units}$ is given by $......$

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