If one diagonal of a square is along the line 8x-15y = 0 and one of its vertex is at (1, 2), then find the equation of sides of the square passing through this vertex.

Let ABCD be the given square and the coordinates of the vertex D be (1, 2).

We are required to find the equations of its sides DC and AD.

Given that BD is along the line 8x − 15y = 0, so its slope is 8/15.

The angles made by BD with sides AD and DC are 45°. Let the slope of DC be m.

Then, tan 45° = (m − 8/15) / (1 + (8m/15))

⇒ 15 + 8m = 15m − 8

⇒ 7m = 23, hence m = 23/7.

Therefore, the equation of side DC is given by:

y − 2 = (23/7)(x − 1)

⇒ 23x − 7y − 9 = 0.

Similarly, the equation of another side AD is given by:

y − 2 = −(7/23)(x − 1)

⇒ 7x + 23y − 53 = 0.

Hence, the equations of the sides are:

23x − 7y − 9 = 0 and 7x + 23y − 53 = 0.