The equation of one diagonal of a square is $2x+3y=5$.. The equation of one diagonal of a square is $2x+3y=5$ and the coordinates of one vertex is $(1,-3)$. Find the equations of two sides of the square which pass through the given vertex.
My Attempt:
Given,
$$2x+3y=5$$
Slope of given line is $-2/3$.
And the slopr of required line be $m$.
A quick thought strikes of a formula
$$\tan \theta=\pm \dfrac {m_1 - m_2}{1+m_1.m_2}$$.
But Is there any other method?
 A: An important property of squares is that their diagonals bisect each other and intersect perpendicularly. 
Note that the point $(1,-3)$ is not on the line $2x + 3y = 5$. But, the diagonal passing through $(1,-3)$ must intersect $2x+3y=5$ perpendicularly. As you've said, the slope of that line is $-\frac{2}{3}$, so the slope of the other diagonal is $\frac{3}{2}$. You can use this to find that the diagonals of the square intersect at $(\frac{37}{13}, -\frac{3}{13})$. Do you have an idea of where to go from there?

A: If $Q(x,y)$ is on line $2x+3y=5$ then rotation around $P$ for $90^{\circ}$ which is point $R(x',y')$ must also lie on this line. So we have also $2x'+3y'=5$.
Now how do we get $R$ if we know $Q$? We go to complex plane. Then $Q = x+yi$ and $P= 1-3i$. Now $PR = i PQ$ and thus 
$$R = P+i(Q-P) = 1-3i+ix-y-i-3= \underbrace{-2-y}_{=x'}+i\underbrace{(x-4)}_{=y'} $$
So we have another equation with $x$ and $y$:
$$ 2(-2-y)+3(x-4)= 5$$
I think you can go on your self here.
A: The diagonals of a square intersect at right angles and bisect the angles at the vertices. Therefore, the direction vectors of the sides are angle bisectors of the diagonals (and so also angle bisectors of their normals).  
We’re given that one diagonal lies on the line $2x+3y=5$. A normal vector to this line can be read directly from the equation: $\mathbf n_1=(2,3)$, and from this we get a normal to the other diagonal, $\mathbf n_2=(3,-2)$. These two vectors have the same length, so their angle bisectors are simply their sum and product. Using the point-normal form of the equation for a line, this gives $$[(3,-2)+(2,3)]\cdot(x,y)-[(3,-2)+(2,3)]\cdot(-1,3)= 5x+y-2=0$$ and $$[(3,-2)-(2,3)]\cdot(x,y)-[(3,-2)-(2,3)]\cdot(-1,3)=x-5y-16=0$$ for equations of the lines on which the sides of the square that meet at the vertex $(1,-3)$ lie.
