Question

Let ABC be a triangle having orthocentre and circumcentre at (9 , 5) and (0 ,0 ) respectively. If the equation of side BC is $2x-y=10$ ,then find the possible coordinates of vertex A.

Answer 1

Co-ordinates of centro-id $\left(G\right)\equiv (\frac{2\times 0+9\times 1}{2+1},\frac{2\times 0+5\times 1}{2+1})\equiv (\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Co-ordinates of centro-id $\left(G\right)\equiv$ $(3,\frac{5}{3})\equiv (\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Now,

$\Rightarrow$ $\frac{x_1+x_2+x_3}{3}=3$

$\Rightarrow$ $x_1+x_2+x_3=9$ ----- ( 1 )

$\Rightarrow$ $\frac{y_1+y_2+y_3}{3}=\frac{5}{3}$

$\Rightarrow$ $y_1+y_2+y_3=5$ ----- ( 2 )

$D(h,k)$ lie on the line $BC$, so it will satisfy the equation $2h-k=10$

Now,

Slope of $CD\times$ Slope of $BC=-1$ [ Since, both are perpendicular to each other ]

$\Rightarrow$ $\frac{k-0}{h-0}\times 2=-1$

$\Rightarrow$ $h=-2k$

Substituting value of $h$ in given equation we get,

$\Rightarrow$ $2(-2k)-k=10$

$\Rightarrow$ $-4k-k=10$

$\Rightarrow$ $-5k=10$

$\therefore$ $k=-2$

$\Rightarrow$ $h=-2k=-2(-2)=4$

So, we got $D$ co-ordinates $(4,-2)=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$

$\Rightarrow$ $\frac{x_2+x_3}{2}=4$

$\Rightarrow$ $x_2+x_3=8$

Substituting above in ( 1 ) we get,

$x_1+8=9$

$\therefore$ $x_1=1$

$\Rightarrow$ $\frac{y_2+y_3}{2}=-2$

$\Rightarrow$ $y_2+y_3=-4$

Substituting above in ( 2 ) we get,

$y_1-4=5$

$\therefore$ $y_1=9$

$\therefore$ The possible co-ordinates of $A$ is $(1,9)$