Let A=(1, 2),B=(2, 1),and C=(-1, -1) be three | vedclass

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

Question

(D) Let $P = (x, y)$. The area of quadrilateral $PABC$ is given by the shoelace formula: $	ext{Area}(PABC) = \frac{1}{2} |(x(2) + 1(1) + 2(-1) + (-1)

Vedclass · Accepted Answer

$3x^2-10xy+3y^2-2x+14y-7=0$

Vedclass · Answer

(D) Let $P = (x, y)$. The area of quadrilateral $PABC$ is given by the shoelace formula: $	ext{Area}(PABC) = \frac{1}{2} |(x(2) + 1(1) + 2(-1) + (-1)y) - (y(1) + 2(2) + 1(-1) + (-1)x)| = \frac{1}{2} |(2x + 1 - 2 - y) - (y + 4 - 1 - x)| = \frac{1}{2} |3x - 2y - 4|$.The area of triangle $PAB$ is given by: $	ext{Area}(	riangle PAB) = \frac{1}{2} |x(2-1) + 1(1-y) + 2(y-2)| = \frac{1}{2} |x + 1 - y + 2y - 4| = \frac{1}{2} |x + y - 3|$.Given that $	ext{Area}(PABC) = 2 	imes 	ext{Area}(	riangle PAB)$,we have:$\frac{1}{2} |3x - 2y - 4| = 2 	imes \frac{1}{2} |x + y - 3|$$|3x - 2y - 4| = 2|x + y - 3|$Squaring both sides:$(3x - 2y - 4)^2 = 4(x + y - 3)^2$$9x^2 + 4y^2 + 16 - 12xy - 24x + 16y = 4(x^2 + y^2 + 9 + 2xy - 6x - 6y)$$9x^2 + 4y^2 + 16 - 12xy - 24x + 16y = 4x^2 + 4y^2 + 36 + 8xy - 24x - 24y$$5x^2 - 20xy + 40y - 20 = 0$Dividing by $5$,we get $x^2 - 4xy + 8y - 4 = 0$.

Let $A=(1, 2)$,$B=(2, 1)$,and $C=(-1, -1)$ be three points. If $P(x, y)$ is a point such that the area of the quadrilateral $PABC$ is twice the area of the triangle $PAB$,then the equation of the locus of $P$ is:

Similar Questions

If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7, \theta \in \left(0, \frac{\pi}{2}\right)$ between the coordinate axes,then $\alpha$ is equal to

If the equation to the locus of points equidistant from the points $(-2, 3)$ and $(6, -5)$ is $a x + b y + c = 0$,where $a > 0$,then the ascending order of $a, b, c$ is

The area of the triangle formed by the intersection of a line parallel to $X$-axis and passing through $P(h, k)$ with the lines $y=x$ and $x+y=2$ is $h^{2}$. The locus of the point $P$ is

If a point $P$ on the line $3x + 5y = 15$ is equidistant from the coordinate axes,then $P$ lies

$A$ straight line $L$ cuts both the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$. The segment of $L$ between the two lines is bisected at the point $(1, 5)$. The equation of $L$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module