Let C be the centroid of the triangle with vertices (3,-1), (1,3) and (2,4).

Question

Let $C$ be the centroid of the triangle with vertices $(3,-1),\ (1,3)$ and $(2,4)$. Let $P$ be the point of intersection of the lines $x+3y-1=0$ and $3x-y+1=0$. Then the line passing through the points $C$ and $P$ also passes through the point:

Accepted Answer

$$C = \left(\frac{3+1+2}3,\frac{-1+3+4}3\right)=(2,2)$$

Solving $x+3y=1$ and $3x-y=-1$: from the second, $y=3x+1$; substitute: $x+3(3x+1)=1 \implies 10x=-2\implies x=-\dfrac15,\ y=\dfrac25$. So $P=\left(-\dfrac15,\dfrac25\right)$.

Line $CP$: slope $=\dfrac{2-\frac25}{2+\frac15}=\dfrac{8/5}{11/5}=\dfrac8{11}$: $y-2=\dfrac8{11}(x-2) \implies 11y-22=8x-16 \implies 8x-11y+6=0$

Check $(-9,-6)$: $8(-9)-11(-6)+6 = -72+66+6=0$

Solutions · 1