If A(1,3) and C(7,5) are two opposite vertice | vedclass

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(D) Let $ABCD$ be a square where $A \equiv (1,3)$ and $C \equiv (7,5)$.Diagonal $AC$ has slope $m_{AC} = \frac{5-3}{7-1} = \frac{2}{6} = \frac{1}{3}$.

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$2x-y+1=0$

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(D) Let $ABCD$ be a square where $A \equiv (1,3)$ and $C \equiv (7,5)$.Diagonal $AC$ has slope $m_{AC} = \frac{5-3}{7-1} = \frac{2}{6} = \frac{1}{3}$.Let the slope of the side $AB$ be $m$.Since the diagonal of a square makes an angle of $45^{\circ}$ with the sides,we have $\left| \frac{m - 1/3}{1 + m(1/3)} ight| = 	an 45^{\circ} = 1$.$\left| \frac{3m-1}{3+m} ight| = 1$.Case $1$: $\frac{3m-1}{3+m} = 1$ $\Rightarrow 3m-1 = 3+m$ $\Rightarrow 2m = 4$ $\Rightarrow m = 2$.The equation of the side passing through $A(1,3)$ with slope $m=2$ is $y-3 = 2(x-1)$ $\Rightarrow y-3 = 2x-2$ $\Rightarrow 2x-y+1 = 0$.Case $2$: $\frac{3m-1}{3+m} = -1$ $\Rightarrow 3m-1 = -3-m$ $\Rightarrow 4m = -2$ $\Rightarrow m = -1/2$.The equation of the side passing through $A(1,3)$ with slope $m=-1/2$ is $y-3 = -1/2(x-1)$ $\Rightarrow 2y-6 = -x+1$ $\Rightarrow x+2y-7 = 0$.Comparing with the options,$2x-y+1=0$ is present.

If $A(1,3)$ and $C(7,5)$ are two opposite vertices of a square,then find the equation of a side passing through $A$.

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