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

Let $ABCD$ be a square where $A\equiv \left(1,3\right)$ & $C\equiv \left(7,5\right).$

$\therefore$Equation of diagonal $AC$ is

$\frac{y-5}{3-5}=\frac{x-7}{1-7}$

$\Rightarrow 3y-15=x-7$

$\Rightarrow 3y-x-8=0 ...\left(1\right)$

$\therefore$Slope of $AC$ is $\frac{1}{3}$

Suppose slope of $AB$is $m.$

$\because$Diagonal makes $45°$with the side.

$\therefore$$\left|\frac{m-{\displaystyle \frac{1}{3}}}{1+{\displaystyle \frac{1}{3}}\times m}\right|=\tan 45°$

$\Rightarrow \left|\frac{3m-1}{3+m}\right|=1$

$\Rightarrow {\left(\frac{3m-1}{3+m}\right)}^2=1$

$\Rightarrow 9m^2-6m+1=9+m^2+6m$

$\Rightarrow 8m^2-12m-8=0$

$\Rightarrow 2m^2-3m-2=0$

$\Rightarrow 2m^2-4m+m-2=0$

$\Rightarrow 2m\left(m-2\right)+1\left(m-2\right)=0$

$\therefore m=-\frac{1}{2}$or $m=2$

when $m=-\frac{1}{2}$

Equation of side passing through $A$ is $y-3=-\frac{1}{2}\left(x-1\right)$

$\Rightarrow 2y-6=-x+1$

$\therefore x+2y-7=0 ...\left(2\right)$

When $m=2,$equation of side passing through $A$ is

$y-3=2\left(x-1\right)$

$\Rightarrow 2x-y+1=0 ...\left(3\right)$

So, option $4$ is correct.