If A(cos alpha, sin alpha),B(sin alpha, -cos | vedclass

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

Question

(B) Let $(h, k)$ be the centroid of the triangle. The coordinates of the centroid are given by the average of the vertices:$h = \frac{\cos \alpha + \s

Vedclass · Accepted Answer

$3(x^2 + y^2) - 2x - 4y + 1 = 0$

Vedclass · Answer

(B) Let $(h, k)$ be the centroid of the triangle. The coordinates of the centroid are given by the average of the vertices:$h = \frac{\cos \alpha + \sin \alpha + 1}{3}$ and $k = \frac{\sin \alpha - \cos \alpha + 2}{3}$Rearranging the terms,we get:$3h - 1 = \cos \alpha + \sin \alpha$ and $3k - 2 = \sin \alpha - \cos \alpha$Squaring both equations and adding them:$(3h - 1)^2 + (3k - 2)^2 = (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2$$(9h^2 - 6h + 1) + (9k^2 - 12k + 4) = (\cos^2 \alpha + \sin^2 \alpha + 2\sin \alpha \cos \alpha) + (\sin^2 \alpha + \cos^2 \alpha - 2\sin \alpha \cos \alpha)$$9h^2 + 9k^2 - 6h - 12k + 5 = 1 + 1 = 2$$9h^2 + 9k^2 - 6h - 12k + 3 = 0$Dividing by $3$,we get:$3(h^2 + k^2) - 2h - 4k + 1 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $3(x^2 + y^2) - 2x - 4y + 1 = 0$.

If $A(\cos \alpha, \sin \alpha)$,$B(\sin \alpha, -\cos \alpha)$,and $C(1, 2)$ are the vertices of a $\Delta ABC$,then as $\alpha$ varies,the locus of its centroid is

Similar Questions

The locus of a point,which moves such that the sum of squares of its distances from the points $(0,0), (1,0), (0,1), (1,1)$ is $18$ units,is a circle of diameter $d$. Then $d^{2}$ is equal to ...... .

$A$ circle touches the $x$-axis and cuts off a chord of length $2l$ from the $y$-axis. The locus of the centre of the circle is

The line segment joining the points $(-5, 1)$ and $(3, 2)$ subtends a right angle at a variable point $P$. Find the locus of the point $P$.

The locus of the point of intersection of the straight lines $tx - 2y - 3t = 0$ and $x - 2ty + 3 = 0$ $(t \in R)$ is

If the coordinates of a point are given by the equations $x = a(1 - \cos \theta )$ and $y = a\sin \theta $,then the locus of the point will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module