Question

The centroid of the triangle whose vertices are (1, 3) (2, 7) and (12, -16) is

• A. (5, -2)
• B. (5, 2)
• C. (-5, 2)
• D. (-5, -2)

Answer 1

The correct option is A (5, -2)

The given coordinates of the vertices of a triangle are:
A(1, 3), B(2, 7) and C(12, -16)
Let point P be the mid point of AB.
$MidpointofAB=(\frac{x1+x2}{2},\frac{y1+y2}{2})$
$=(\frac{1+2}{2},\frac{3+7}{2})$
$=(\frac{3}{2},5)$
Coordinates of point P = $=(\frac{3}{2},5)$
Let PC be the median of the triangle ABC.
Centroid G divides the median in the ratio 2:1

The equation for the point that divides a line segment joining the points $(x_1,y_1)and(x_2,y_2)$ in the ratio m : n is: $(\frac{(n\times x_1+m\times x_2)}{m+n},\frac{n\times y_1+m\times y_2}{m+n})$

Here,
$(x_1,y_1)=(12,-16)\text{and}(x_2,y_2)=(1.5,5)$

Applying the formula, we get
$=(\frac{2\left(1.5\right)+1\left(12\right)}{3},\frac{2\left(5\right)+1(-16)}{3})$
$=(\frac{15}{3},\frac{-6}{3})$
$=(5,\frac{-6}{3})$$=(5,-2)$