IIT-JEE 1978Question 1
A triangle $ABC$ has sides $AB=AC=5$ cm and $BC=6$ cm. Triangle $A'B'C'$ is the reflection of the triangle $ABC$ in a line parallel to $AB$ placed at a distance $2$ cm from $AB$, outside the triangle $ABC$. Triangle $A''B''C''$ is the reflection of the triangle $A'B'C'$ in a line parallel to $B'C'$ placed at a distance of $2$ cm from $B'C'$ outside the triangle $A'B'C'$. Find the distance between $A$ and $A''$.
Points and Straight Lines
Answer$8\sqrt{\dfrac{17}{5}}$
Solution
Take $B(-3,0)$, $C(3,0)$, $A(0,4)$; reflecting in $4x-3y+22=0$ gives $A'$, $B'$, $C'$, and reflecting $A'$ in the line parallel to $B'C'$ at distance $2$ outside gives $A''$, with $AA''=\frac{8\sqrt{85}}{5}=8\sqrt{\frac{17}{5}}$.