Question

In triangle ABC, base BC and area of triangle are fixed. The locus of the centriod of triangle ABC is a straight line that is:

• A. parallel to side BC
• B. right bisector of side BC
• C. right angle of BC
• D. inclined at an angle ${\sin }^{-1}\left(\sqrt{\Delta }/BC\right)$ to side BC

Answer 1

The correct option is A parallel to side BC

Let the base $BC=b$
Now base is fixed and area is fixed.
Hence altitude of the triangle is fixed since area is $=\frac{1}{2}b.h$.
Now let $B=(x_1,0)$ $C=(x_2,0)$ and $h$ be denoted by the magnitude of $(0,y_1)$.

Then the third vertex will be $(x_i,y_1)$ where $x_iϵR$.

Hence $G=(\frac{x_i+x_2+x_1}{3},\frac{y_1}{3})$.

Now area and base are fixed.
So height is fixed.
Hence y-coordinate of G is fixed.
Also base is fixed, hence G varies as $x_i$ varies.
Hence the locus of centro-id is a straight line parallel to the Base of the triangle.