Question

If the vertices $A$ and $B$ of a $△ABC$ are given by $(2,5)$ and $(4,-11)$ respectively and $C$ moves along the line $L=9x+7y+4=0$ then the locus of the centroid of the $△ABC$ is

• A. a circle
• B. any straight line
• C. a line parallel to $L$
• D. a line perpendicular to $L$

Answer 1

The correct option is C a line parallel to $L$

Given that:

$A(2,5),B(4,-11)$ and $C$ moves on line $L=9x+7y+4=0$

To find:

Locus of centroid of the $△ABC$

Solution:

Let coordinate of $C(x,y)$ which satisfies the line $L.$

And Coordinates of centriod $(h,k)$

$h=\frac{2+4+x}{3}$

or, $3h=6+x$

or, $x=3h-6$ $..............\left(i\right)$

$k=\frac{5-11+y}{3}$

or, $3k=-6+y$

or, $y=3k+6$ $.............\left(ii\right)$

Putting value of $x$ and $y$ from eqn.(i) and (ii) in line $L$ we get,

$9(3h-6)+7(3k+6)+4=0$

or, $27h-54+21k+42+4=0$

or, $27h+21k-8=0$

Now, Locus of Centroid is $27x+21y-8=0.$

$\to y=\frac{8}{21}-\frac{9}{7}x$

slope is $m=-\frac{9}{7}$ exactly as the slope of line $L$ thus we can say that the locus is parallel to the line $L$.

Hence, C is the correct option.