The coordinates of the vertices P, Q, R and S | vedclass

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(D) Let the side length of the square be $a$. Since $P$ and $Q$ lie on $AB$ (the $x$-axis),let $P = (x_1, 0)$ and $Q = (x_1 + a, 0)$.Then $S = (x_1, a

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

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(D) Let the side length of the square be $a$. Since $P$ and $Q$ lie on $AB$ (the $x$-axis),let $P = (x_1, 0)$ and $Q = (x_1 + a, 0)$.Then $S = (x_1, a)$ and $R = (x_1 + a, a)$.Point $S$ lies on $AC$. The equation of line $AC$ passing through $(0, 0)$ and $(2, 1)$ is $y = \frac{1}{2}x$. Substituting $S(x_1, a)$,we get $a = \frac{1}{2}x_1 \Rightarrow x_1 = 2a$.Point $R$ lies on $BC$. The equation of line $BC$ passing through $(3, 0)$ and $(2, 1)$ is $y - 0 = \frac{1-0}{2-3}(x - 3)$ $\Rightarrow y = -(x - 3)$ $\Rightarrow x + y = 3$.Substituting $R(x_1 + a, a)$,we get $(x_1 + a) + a = 3 \Rightarrow x_1 + 2a = 3$.Substituting $x_1 = 2a$ into the second equation: $2a + 2a = 3$ $\Rightarrow 4a = 3$ $\Rightarrow a = \frac{3}{4}$.Then $x_1 = 2(\frac{3}{4}) = \frac{3}{2}$.The coordinates are $P(\frac{3}{2}, 0)$,$Q(\frac{3}{2} + \frac{3}{4}, 0) = (\frac{9}{4}, 0)$,$R(\frac{9}{4}, \frac{3}{4})$,and $S(\frac{3}{2}, \frac{3}{4})$.

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

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