Let A(-3,2) and B(-2,1) be the vertices of a triangle ABC.

Question

Let $A(-3,2)$ and $B(-2,1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x+4y+2=0$, then the vertex $C$ lies on the line:

Accepted Answer

Let $C=(x_1,y_1)$. Centroid $=\left(\dfrac{x_1-5}3,\dfrac{y_1+3}3\right)$, on $3x+4y+2=0$:
$$3\left(\frac{x_1-5}3\right)+4\left(\frac{y_1+3}3\right)+2=0 \implies (x_1-5)+\frac{4(y_1+3)}3+2=0$$
$$3x_1-15+4y_1+12+6=0 \implies 3x_1+4y_1+3=0$$

Vertex $C$ lies on $3x+4y+3=0$.

Solutions · 1