Question

Let ABC be a triangle having orthocentre & 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.

• A. $(1,9)$
• B. $(4,3)$
• C. $(2,15)$
• D. $(3,15)$

Answer 1

The correct option is A $(1,9)$

$\begin{array}{c}Q\equiv \left(\frac{2\times 0+9\times 1}{2+1},\frac{2\times 0+5\times 1}{2+1}\right)\equiv \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\\ Q\equiv \left(3,\frac{5}{3}\right)\equiv \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\\ x_1+x_2+x_3=9\\ y_1+y_2+y_3=5\\ LetD\left(h,k\right)lieonlineBC,\\ 2h-k=10\\ slopeofCD\times slopeofBC=-1\\ \Rightarrow \frac{k-0}{h-0}\times 2=-1\\ \Rightarrow h=-2k\\ and,\\ 2\times -2k-k=10\\ k=2\\ h=2k=4\\ D\equiv \left(4,-2\right)\equiv \left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)\\ \Rightarrow x_2+x_3=8and{x}_1+x_2+x_3=9\\ and,y_2+y_3=-4and{y}_1+y_2+y_3=5\\ \begin{array}{c}{x}_1=1\\ y_1=9\end{array}\}\\ Hence,thepossiblepointofAis\left(1,9\right).\end{array}$

Option $A$ is the correct answer.