The distance of the origin from the centroid of the triangle whose two sides have the equa…

Question

The distance of the origin from the centroid of the triangle whose two sides have the equations $x-2y+1=0$ and $2x-y-1=0$ and whose orthocenter is $\left(\dfrac73,\dfrac73\right)$ is:

Accepted Answer

Vertex $A$ = intersection of $x-2y+1=0$ and $2x-y-1=0$: solving gives $A=(1,1)$.

Altitude from $B$ perpendicular to $AC$ (i.e. perpendicular to $x-2y+1=0$, slope $\dfrac12$, so altitude slope $-2$): $2x+y+\lambda=0$ through $H\left(\dfrac73,\dfrac73\right)$: $\dfrac{14}3+\dfrac73+\lambda=0\implies\lambda=-7$, so altitude: $2x+y-7=0$.

Solving with $2x-y-1=0$ (line $AC$... actually $x-2y+1=0$): $x=2,\ y=3$, giving $B=(2,3)$.

Altitude from $C$ perpendicular to $AB$: $x+2y+\lambda=0$ through $H$: $\dfrac73+\dfrac{14}3+\lambda=0\implies\lambda=-7$, so $x+2y-7=0$.

Solving with $2x-y-1=0$: $x=3,\ y=2$, giving $C=(3,2)$.

Centroid $G = \left(\dfrac{1+2+3}3,\dfrac{1+3+2}3\right)=(2,2)$.

$$OG = \sqrt{4+4}=2\sqrt2$$

Solutions · 1