The points (1,3) and (5,1) are two opposite v | vedclass

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(A) Let the opposite vertices be $A(1,3)$ and $C(5,1)$. The midpoint of $AC$ is $(\frac{1+5}{2}, \frac{3+1}{2}) = (3,2)$.Since the diagonals of a rect

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$(4,4), (2,0)$

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(A) Let the opposite vertices be $A(1,3)$ and $C(5,1)$. The midpoint of $AC$ is $(\frac{1+5}{2}, \frac{3+1}{2}) = (3,2)$.Since the diagonals of a rectangle bisect each other,the midpoint of the other diagonal $BD$ must also be $(3,2)$.Let the vertices $B$ and $D$ be $(x_1, y_1)$ and $(x_2, y_2)$. Since they lie on $y = 2x + c$,we have $y_1 = 2x_1 + c$ and $y_2 = 2x_2 + c$.The midpoint is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (3,2)$,so $x_1+x_2 = 6$ and $y_1+y_2 = 4$.Substituting $y_1, y_2$: $(2x_1+c) + (2x_2+c) = 4 \implies 2(x_1+x_2) + 2c = 4 \implies 2(6) + 2c = 4 \implies 12 + 2c = 4 \implies c = -4$.Thus,the line is $y = 2x - 4$.Also,the vector $\vec{AB} = (x_1-1, y_1-3)$ must be perpendicular to $\vec{BC} = (5-x_1, 1-y_1)$.Using the dot product: $(x_1-1)(5-x_1) + (2x_1-4-3)(1-(2x_1-4)) = 0$.$(x_1-1)(5-x_1) + (2x_1-7)(5-2x_1) = 0$.$-x_1^2 + 6x_1 - 5 - 4x_1^2 + 24x_1 - 35 = 0 \implies -5x_1^2 + 30x_1 - 40 = 0 \implies x_1^2 - 6x_1 + 8 = 0$.$(x_1-4)(x_1-2) = 0$,so $x_1 = 4$ or $x_1 = 2$.If $x_1 = 4, y_1 = 2(4)-4 = 4$. If $x_1 = 2, y_1 = 2(2)-4 = 0$.The vertices are $(4,4)$ and $(2,0)$.

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

The points $(1,3)$ and $(5,1)$ are two opposite vertices of a rectangle. The other two vertices lie on the line $y = 2x + c$,where $c$ is a constant. The coordinates of the other two vertices are:

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