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

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(C) Diagonals of a rectangle bisect each other.Midpoint of $(1,3)$ and $(5,1)$ is $(\frac{1+5}{2}, \frac{3+1}{2}) = (3,2)$.Since the other two vertice

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$(2,0)$

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(C) Diagonals of a rectangle bisect each other.Midpoint of $(1,3)$ and $(5,1)$ is $(\frac{1+5}{2}, \frac{3+1}{2}) = (3,2)$.Since the other two vertices lie on the line $y=2x+c$,the midpoint $(3,2)$ must also lie on this line.Substituting $(3,2)$ into $y=2x+c$ gives $2 = 2(3) + c$,so $c = -4$.The equation of the line containing the other two vertices is $y=2x-4$.Let the coordinates of a vertex be $(x, 2x-4)$.Since the angle between adjacent sides of a rectangle is $90^{\circ}$,the product of the slopes of the sides meeting at a vertex is $-1$.Let the given vertices be $A(1,3)$ and $C(5,1)$. Let the other vertices be $B(x, 2x-4)$ and $D$. The slope of $AB$ is $\frac{2x-4-3}{x-1} = \frac{2x-7}{x-1}$.The slope of $BC$ is $\frac{2x-4-1}{x-5} = \frac{2x-5}{x-5}$.Since $AB \perp BC$,we have $(\frac{2x-7}{x-1})(\frac{2x-5}{x-5}) = -1$.$(2x-7)(2x-5) = -(x-1)(x-5)$$4x^2 - 24x + 35 = -(x^2 - 6x + 5)$$4x^2 - 24x + 35 = -x^2 + 6x - 5$$5x^2 - 30x + 40 = 0$$x^2 - 6x + 8 = 0$$(x-4)(x-2) = 0$So $x=4$ or $x=2$.If $x=4$,$y=2(4)-4=4$. If $x=2$,$y=2(2)-4=0$.The vertices are $(4,4)$ and $(2,0)$.Thus,one of the vertices is $(2,0)$.

The points $(1,3)$ and $(5,1)$ are opposite vertices of a diagonal of a rectangle. If the other two vertices lie on the line $y=2x+c$,then one of the vertices on the other diagonal is

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