Question

If the circumcentre of a triangle lies at the origin and the centroid is the middle point of the line joining the points $(a^2+1,a^2+1)$ and $(2a,-2a)$, then find the locus of orthocenter.

Answer 1

We have circumcenter of triangle lies at the origin, so the coordinates of circumcenter are $(0,0)$

Now, let $A(a^2+1,a^2+1)$ and $B(2a,-2a)$ are the given points.

Now, let $P(m,n)$ be the mid point of $AB$.

Then,

$m=\frac{a^2+1+2a}{2}=\frac{(a+1{)}^2}{2}$

$n=\frac{a^2+1-2a}{2}=\frac{(a-1{)}^2}{2}$

So, cordinates of the centroid of the triangle are $(\frac{(a+1{)}^2}{2},\frac{(a-1{)}^2}{2})$

Now, we know that circumcenter, orthocenter and centriod of a triangle lie on a line.

So, orthocentre will lie on the line joining circumcenter and the centroid.

Now, we know that equation pf the line joining points $(x_1,y_1)$ and $x_2,y_2)$ is $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

So, the equation of a line joining $(0,0)$ and $(\frac{(a+1{)}^2}{2},\frac{(a-1{)}^2}{2})$ is

$\frac{y-0}{x-0}=\frac{\frac{(a-1{)}^2}{2}-0}{\frac{(a+1{)}^2}{2}-0}$

$\Rightarrow$ $\frac{y}{x}=\frac{(a-1{)}^2}{(a+1{)}^2}$

$\Rightarrow$ $(a-1{)}^{2}x-(a+1{)}^{2}y=0$