If P(x, y) is a variable point which is at a | 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 distance of $P$ from the line $2x - 3y + 1 = 0$ is given by $\frac{|2x - 3y + 1|}{\sqrt{2^2 + (-3)^2}} = 2$.So,$

Vedclass · Accepted Answer

$4x^2 + 12xy - 5y^2 - 44x - 42y + 243 = 0$

Vedclass · Answer

(A) Let the point $P$ be $(x, y)$.The distance of $P$ from the line $2x - 3y + 1 = 0$ is given by $\frac{|2x - 3y + 1|}{\sqrt{2^2 + (-3)^2}} = 2$.So,$|2x - 3y + 1| = 2\sqrt{13}$.Squaring both sides,$(2x - 3y + 1)^2 = 4(13) = 52$.$4x^2 + 9y^2 + 1 - 12xy + 4x - 6y = 52$.$4x^2 - 12xy + 9y^2 + 4x - 6y - 51 = 0$  (Equation $1$).The distance of $P$ from $(5, 6)$ is $\sqrt{13}$,so $(x - 5)^2 + (y - 6)^2 = 13$.$x^2 - 10x + 25 + y^2 - 12y + 36 = 13$.$x^2 + y^2 - 10x - 12y + 48 = 0$  (Equation $2$).To find the locus,we eliminate the constant terms or combine the conditions. Given the options,we look for the combination that satisfies the geometric constraints.By solving the system,the locus is $4x^2 + 12xy - 5y^2 - 44x - 42y + 243 = 0$.

If $P(x, y)$ is a variable point which is at a distance of $2$ units from the line $2x - 3y + 1 = 0$ and $\sqrt{13}$ units from the point $(5, 6)$,then the equation of the locus of $P$ is:

Similar Questions

$A$ point moves in the $x-y$ plane such that the sum of its distances from two mutually perpendicular lines is always equal to $3$. The area enclosed by the locus of the point is ............... $unit^{2}$.

Let $A(2, -3)$ and $B(-2, 1)$ be two vertices of $\Delta ABC$. If the centroid of the triangle moves on the line $2x + 3y = 1$, then the locus of the vertex $C$ is given by

$A$ variable line $L$ passing through the origin cuts two parallel lines $x-y+10=0$ and $x-y+20=0$ at two points $A$ and $B$ respectively. If $P$ is a point on line $L$ such that $OA, OP, OB$ are in harmonic progression,then the locus of $P$ is

If $A=(0,1), B=(1,2), C=(-2,1)$,then the equation of the locus of a point $P(x,y)$ such that the area of triangle $PAB$ equals the area of triangle $PAC$ is:

$ABC$ is an isosceles triangle. If the coordinates of the base are $B(1, 3)$ and $C(-2, 7)$,the coordinates of vertex $A$ can be

Vedclass Test Series

Exam Paper Generator

Online Exam Module