Question

The co-ordinates of the vertices $P,Q,R$ & $S$ of square $PQRS$ inscribed in the $△ABC$ with vertices $A\equiv (0,0),B\equiv (3,0)$ & $C\equiv (2,1)$ given that two of its vertices $P,Q$ are on the side $AB$ are respectively

• A. $(\frac{1}{4},0),(\frac{3}{8},0),(\frac{3}{8},\frac{1}{8})$ & $(\frac{1}{4},\frac{1}{8})$
• B. $(\frac{1}{2},0),(\frac{3}{4},0),(\frac{3}{4},\frac{1}{4})$ & $(\frac{1}{2},\frac{1}{4})$
• C. $(1,0),(\frac{3}{2},0),(\frac{3}{2},\frac{1}{2})$ & $(1,\frac{1}{2})$
• D. $(\frac{3}{2},0),(\frac{9}{4},0),(\frac{9}{4},\frac{3}{4})$ & $(\frac{3}{2},\frac{3}{4})$

Answer 1

The correct option is D $(\frac{3}{2},0),(\frac{9}{4},0),(\frac{9}{4},\frac{3}{4})$ & $(\frac{3}{2},\frac{3}{4})$

Let co-ordinates of P are $(x_1,0)$ and side of square is 'a'
$\therefore Q(x_1+a,0)$
$S(x_1,a)$
$R(x_1+a,a)$
Now,
$m_{AS}=m_{AC}$
$\Rightarrow \frac{a}{x_1}=\frac{1}{2}\Rightarrow x_1=2a$ ......(1)
$m_{BR}=m_{BC}$
$\Rightarrow \frac{a}{x_1+a-3}=-1\Rightarrow x_1+2a-3=0$ ......(2)
from (1) & (2) $a=\frac{3}{4}$ & $x_1=\frac{3}{2}$
Hence co-ordinates of P, Q, R, & S can be determined.

as $(\frac{3}{2},0),(\frac{9}{4},0),(\frac{9}{4},\frac{3}{4})$ and $(\frac{3}{2},\frac{3}{4})$