Question

If a vertex of a triangle is (1, 1) and the mid-points of the two sides through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is given as (x, y). Find x+3y.

• A. 8

Answer 1

The correct option is A 8

Let the coordinates of B and C be $(x_B,y_B)\text{and}(x_C,y_C)$ respectively.

We know that mid-point of the line joining two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(\frac{x_B+1}{2},\frac{y_B+1}{2})=(-1,2)\frac{x_B+1}{2}=-1;\frac{y_B+1}{2}=2\therefore x_B=-3;y_B=3$
Coordinates of B is $(-3,3).$

Similarly we can find coordinates of C as (5,3).

Hence, the centroid of triangle, G is $(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3})$
$G(\frac{1-3+5}{3},\frac{1+3+3}{3})G(1,\frac{7}{3})$
$\therefore$ The centroid of the triangle is $(1,\frac{7}{3})$

Here, x = 1 and y = $\frac{7}{3}$
x+3y = 1 + 7 = 8