The side $B C$ of a triangle $A B C$ lies on $x$-axis. If its circumcentre is $(5,7)$ and orthocentr

Question

The side $B C$ of a triangle $A B C$ lies on $x$-axis. If its circumcentre is $(5,7)$ and orthocentre is $(2,3)$ and vertices $A=\left(a_1, b_1\right), B=\left(a_2, b_2\right), C=\left(a_3, b_3\right) \quad$ (where $\left.a_2<a_3\right)$. Then

Professor PC Thomas Classes · Answer

$\begin{aligned} & H^1 S=R=\sqrt{109} \ & G=\left(4, \frac{17}{3}ight) \ & \frac{a+b+c}{3}=4 \Rightarrow a+b+c=12 \ & \frac{d}{3}=\frac{17}{3} \Rightarrow d=17 \ & S B=\sqrt{109} \Rightarrow a=5 \pm \sqrt{60} \ & 	herefore B=(5-\sqrt{60}, 0), C=(5+\sqrt{60}, 0) \ & a+b+c=4 	imes 3=12 \ & 10+C=12 \ & C=2\end{aligned}$