The locus of a point which moves such that th | vedclass

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

Question

(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Let the moving point be $P(x, y)$.The sum of the squares of the

Vedclass · Accepted Answer

Centroid of the triangle

Vedclass · Answer

(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Let the moving point be $P(x, y)$.The sum of the squares of the distances from $P$ to the vertices is constant,say $K$:$(x - x_1)^2 + (y - y_1)^2 + (x - x_2)^2 + (y - y_2)^2 + (x - x_3)^2 + (y - y_3)^2 = K$Expanding the terms:$3x^2 - 2x(x_1 + x_2 + x_3) + 3y^2 - 2y(y_1 + y_2 + y_3) + (x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2) = K$Dividing by $3$:$x^2 - 2x\left(\frac{x_1 + x_2 + x_3}{3}ight) + y^2 - 2y\left(\frac{y_1 + y_2 + y_3}{3}ight) = 	ext{constant}$This is the equation of a circle with centre at $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}ight)$,which is the centroid of the triangle.

The locus of a point which moves such that the sum of the squares of its distances from the three vertices of a triangle is constant,is a circle whose centre is at the

Similar Questions

Three distinct points $A, B$,and $C$ are given in a two-dimensional coordinate plane such that for each point,the ratio of its distance from $(1, 0)$ to its distance from $(-1, 0)$ is equal to $\frac{1}{3}$. What is the circumcenter of triangle $ABC$?

The locus of a point,such that the difference of the squares of the lengths of the tangents drawn from it to two given circles is constant,is:

The locus of the point of intersection of tangents to the circle $x = a \cos \theta, y = a \sin \theta$ at the points whose parametric angles differ by $\pi / 2$ is:

If a circle of constant radius $3k$ passes through the origin $O$ and meets the coordinate axes at $A$ and $B$,then the locus of the centroid of the triangle $OAB$ is

The angle between a pair of tangents drawn from a point $P$ to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$ is $2\alpha$. The equation of the locus of the point $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module