The locus of P such that the area of Delta PA | vedclass

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

Question

(B) Let $P$ be $(x, y)$. The area of $\Delta PAB$ 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 the c

Vedclass · Accepted Answer

$(x + 3y + 1)(x + 3y - 23) = 0$

Vedclass · Answer

(B) Let $P$ be $(x, y)$. The area of $\Delta PAB$ 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 the coordinates $P(x, y)$,$A(2, 3)$,and $B(-4, 5)$:$\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 $2$: $x + 3y - 23 = 0$.Therefore,the locus is $(x + 3y + 1)(x + 3y - 23) = 0$.

The locus of $P$ such that the area of $\Delta PAB = 12 \text{ sq. units}$,where $A(2, 3)$ and $B(-4, 5)$ is:

Similar Questions

Let the algebraic sum of the perpendicular distances from the points $A(2, 0)$,$B(0, 2)$,and $C(1, 1)$ to a variable line be zero. Then,all such lines:

Suppose $P$ and $Q$ are the midpoints of the sides $AB$ and $BC$ of a triangle where $A(1, 3)$,$B(3, 7)$,and $C(7, 15)$ are vertices. Then the locus of $R$ satisfying $AC^2 + QR^2 = PR^2$ is

The equation $x^{3}-y x^{2}+x-y=0$ represents

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b, a + b)$,then:

If a point $P$ is equidistant from the points $A(a + b, b - a)$ and $B(a - b, a + b)$,find the locus of $P$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module