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

Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$,then the equation of the locus of the midpoint $P$ of $BC$ is

$A$ variable line passes through a fixed point $(a, b)$ and meets the coordinate axes at $A$ and $B$. The locus of the point of intersection of lines drawn through $A$ and $B$ parallel to the coordinate axes is:

If the locus of the centroid of the triangle with vertices $A(a, 0)$,$B(a \cos t, a \sin t)$ and $C(b \sin t, -b \cos t)$ ($t$ is a parameter) is $9x^2 + 9y^2 - 6x = 49$,then the area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is

If two mutually perpendicular lines through the point $A(1, 1)$ intersect the $x$-axis and $y$-axis at points $B$ and $C$ respectively,then the locus of the centroid of $\Delta ABC$ is -

If the ends of the hypotenuse of a right-angled triangle are $(0, a)$ and $(a, 0)$,then the locus of the third vertex is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module