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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $......$

Similar Questions

$A(-4,0)$ and $B(4,0)$ are two fixed points. $C$ and $D$ are two points on the $Y$-axis such that $CD=4$ and $C$ is a point below $D$. Then the locus of the point of intersection of the lines $AC$ and $BD$ is

$A$ line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is

Given $\frac{x}{a} + \frac{y}{b} = 1$ and $ax + by = 1$ are two variable lines,where $a$ and $b$ are parameters connected by the relation $a^2 + b^2 = ab$. The locus of the point of intersection has the equation:

If $A=(-1, 2)$ and $B=(1, -2)$ are two points and $P$ is a variable point such that the area of $\triangle PAB$ is always $1$,then the equation of the locus of $P$ is

$A$ straight line meets the $X$ and $Y$ axes at the points $A$ and $B$ respectively. If $AB = 6$ units,then the locus of the point $P$ which divides the line segment $AB$ such that $AP : PB = 2 : 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module