(0, -1) and (0, 3) are two opposite vertices | vedclass

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(C) Let the given vertices be $A(0, -1)$ and $C(0, 3)$.The length of the diagonal $AC = \sqrt{(0-0)^2 + (3 - (-1))^2} = \sqrt{0^2 + 4^2} = 4$.Let the

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$(2, 1), (-2, 1)$

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(C) Let the given vertices be $A(0, -1)$ and $C(0, 3)$.The length of the diagonal $AC = \sqrt{(0-0)^2 + (3 - (-1))^2} = \sqrt{0^2 + 4^2} = 4$.Let the other two vertices be $B(x, y)$ and $D(x', y')$.In a square,the diagonals bisect each other at right angles and are equal in length.The midpoint of $AC$ is $(\frac{0+0}{2}, \frac{-1+3}{2}) = (0, 1)$.Since the diagonals bisect each other,the midpoint of $BD$ is also $(0, 1)$.Thus,$\frac{x+x'}{2} = 0 \Rightarrow x' = -x$ and $\frac{y+y'}{2} = 1 \Rightarrow y' = 2-y$.Also,the distance from the midpoint $(0, 1)$ to any vertex is half the diagonal length,which is $2$.For vertex $B(x, y)$,the distance to $(0, 1)$ is $\sqrt{(x-0)^2 + (y-1)^2} = 2$,so $x^2 + (y-1)^2 = 4$.Since $AB \perp BC$,the slope of $AB$ is $\frac{y+1}{x-0} = \frac{y+1}{x}$ and the slope of $BC$ is $\frac{y-3}{x-0} = \frac{y-3}{x}$.Since $AB \perp BC$,their product is $-1$: $(\frac{y+1}{x})(\frac{y-3}{x}) = -1$ $\Rightarrow (y+1)(y-3) = -x^2$ $\Rightarrow y^2 - 2y - 3 = -x^2$ $\Rightarrow x^2 + y^2 - 2y - 3 = 0$.Substituting $x^2 = 4 - (y-1)^2 = 4 - (y^2 - 2y + 1) = 3 + 2y - y^2$ into the equation:$(3 + 2y - y^2) + y^2 - 2y - 3 = 0$,which is $0=0$.This confirms the midpoint logic. Since $AC$ is vertical,$BD$ must be horizontal,so $y = 1$.Substituting $y=1$ into $x^2 + (y-1)^2 = 4$,we get $x^2 + 0 = 4$,so $x = \pm 2$.Thus,the other two vertices are $(2, 1)$ and $(-2, 1)$.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

$(0, -1)$ and $(0, 3)$ are two opposite vertices of a square. The other two vertices are

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