Question

Let $PQRS$ be a square inscribed in the triangle with vertices $A(0,0),B(3,0)$ and $C(2,1)$. Given that $P,Q$ are on the side $AB,R$ on side $BC$ and $S$ on side $AC$.
The centre of square $PQRS$ is

• A. $\left(\frac{13}{8},\frac{8}{3}\right)$
• B. $\left(\frac{15}{8},\frac{3}{8}\right)$
• C. $\left(\frac{15}{8},\frac{8}{3}\right)$
• D. $\left(\frac{14}{3},\frac{8}{3}\right)$

Answer 1

Answer: (B)

Let PQRS be the square inscribed in $△ABC$ with length of each side equal to a. Let the coordinates of $P$ be $(p,0)$. So, the coordinates of $Q,R,S$ are respectively $(p+a,0),(p+a,a)$ and (p, a).
Now, equation of $AC$ is $y=\frac{1}{2}x$ and $S(p,a)$ lies on it.
$\Rightarrow a=\frac{p}{2}\text{ or }p=2a$
Also equation of $BC$ is $(x+y)=3$ and $R(p+a,a)$ lies on it.

$\therefore (p+a)+a=3\Rightarrow 4a=3\Rightarrow a=\frac{3}{4}$ and $p=\frac{3}{2}$
So, $P\left(\frac{3}{2},0\right),Q\left(\frac{9}{4},0\right),R\left(\frac{9}{4},\frac{3}{4}\right)$ and $S\left(\frac{3}{2},\frac{3}{4}\right)$.
Centre of square PQRS
$=\left(\frac{\frac{3}{2}+\frac{9}{4}}{2},\frac{0+\frac{3}{4}}{2}\right)=\left(\frac{15}{8},\frac{3}{8}\right)$