Question

Let P be a point inside a triangle ABC with $\angle$ABC $={90}^o$. Let $P_1$ and $P_2$ be the images of P under reflection in AB and BC respectively. The distance between the circumcentres of triangles ABC and $P_{1}P{P}_2$ is?

• A. $\frac{AB}{2}$
• B. $\frac{AP+BP+CP}{3}$
• C. $\frac{AC}{2}$
• D. $\frac{AB+BC+AC}{2}$

Answer 1

The correct option is C $\frac{AC}{2}$

$AB$ $P_{1}P$

$BC$ $P_{2}P$

$\Delta PP,P_2$

$P{P}_1{P}_2$

$ABCBR=R$

$=AD=DC=\frac{AC}{2}$

$P(x,y),P_1(-x/y)$

$P_2(x,-y)$

$\Delta P{P}_1{P}_2$

$P_1{P}_2=(\frac{-x+x}{2},\frac{y-y}{2})$

$=(0,0)$

$BD=AD=\frac{AC}{2}$