If the locus of the centroid of the triangle | vedclass

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

Question

(B) Let the centroid be $(x, y)$. The coordinates of the centroid are given by: $(x, y) = \left(\frac{a + a \cos t + b \sin t}{3}, \frac{0 + a \sin t

Vedclass · Accepted Answer

$\frac{7}{2}$

Vedclass · Answer

(B) Let the centroid be $(x, y)$. The coordinates of the centroid are given by: $(x, y) = \left(\frac{a + a \cos t + b \sin t}{3}, \frac{0 + a \sin t - b \cos t}{3}ight)$ This implies: $3x = a + a \cos t + b \sin t \Rightarrow 3x - a = a \cos t + b \sin t$ $3y = a \sin t - b \cos t$ Squaring and adding these two equations: $(3x - a)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$ $9x^2 + a^2 - 6ax + 9y^2 = a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t)$ $9x^2 + 9y^2 - 6ax + a^2 = a^2 + b^2$ $9x^2 + 9y^2 - 6ax = b^2$ Comparing this with the given locus $9x^2 + 9y^2 - 6x = 49$,we get $a = 1$ and $b^2 = 49$,so $b = 7$. The line equation is $\frac{x}{1} + \frac{y}{7} = 1$. The intercepts are $x = 1$ and $y = 7$. The area of the triangle formed with the coordinate axes is $\frac{1}{2} 	imes |x_{intercept}| 	imes |y_{intercept}| = \frac{1}{2} 	imes 1 	imes 7 = \frac{7}{2}$.

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

Similar Questions

If $2a + b + 3c = 0$,then the line $ax + by + c = 0$ passes through the fixed point that is

The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is

$A$ line moves such that the portion of it intercepted between the coordinate axes is of constant length $a$. Then,the locus of the midpoint of that line segment is

$A$ line $L$ passing through the point $P(-5, -4)$ cuts the lines $x-y-5=0$ and $x+3y+2=0$ at $Q$ and $R$ respectively such that $\frac{18}{PQ} + \frac{15}{PR} = 2$. Then the slope of the line $L$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module