Question

The line containing the diagonal of a square whose vertices are A(-3, 3), B(3, 3), C(3, -3), and D(-3, -3). Find two equations, one for each diagonal.

Answer 1

Attached to the answer is a graph of the square, based on the points given. From there you can easily guess the equations of the diagonal lines.

However, you do not need a graph in order to solve the problem. 

AC and BD are two pairs of points diagonally across from one another. This is because the pairs do not share the same x or y value. 

Based on the equation of the line:
$y = mx+b$ 
We must find the slope(m) and the y-intercept(b).

Finding slope (m):
Lets use the equation below:
$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Line AC:
$m = \frac{-3-3}{3-(-3)}$
$m = \frac{-6}{6}$
∴ m = -1
∴ $y_{BD} = -1x +b$

Line BD:
$m = \frac{-3-3}{-3-3}$
$m = \frac{-6}{-6}$
∴ m = 1
∴ $y_{BD} = 1x +b$

Find y-intercept (b):
In order to find the y-intercept(b) of the line, we must plug in a point that is on that line.

Line AC:
y = -1x + b
3 = -1(-3) + b
3 = 3 + b
b = 0 

Line BD:
y = 1x + b
3 = 1(3) + b
3 = 3 + b
b = 0

Therefore, the equations for the diagonals are as follows:
$y_{AC} = -x$
$y_{BD} = x$

Hope this helps!