Question

The coordinates of the vertices $P,Q,R$ & $S$ of square $PQRS$ inscribed in the triangle $ABC$ with vertices $A(0,0),B(3,0)$ & $C(2,1)$ given that two of its vertices $P,Q$ are on the side $AB$ are respectively?

Answer 1

Given $PQRS$ is a square = $PQ=QR=RS=SP$

$\frac{SP}{AP}=\frac{1}{2},AP=2SP=2PQ$

$\frac{QR}{QB}=\frac{1}{1},QR=QB=PQ$

As, $AP+PQ+QB=AB=3$

$2PQ+PQ+PQ=3$

$4PQ=3,PQ=\frac{3}{4},QR=RS=SP=\frac{3}{4}$

Hence, $P=(AP,0)=(2\ast \frac{3}{4},0)=(\frac{3}{2},0)$

$Q=(AP+PQ,0)=(\frac{3}{2}+\frac{3}{4},0)=(\frac{9}{4},0)$

$R=(\frac{9}{4},\frac{3}{4})$

$S=(\frac{3}{2},\frac{3}{4})$