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.

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:
[tex]y = mx+b[/tex]
We must find the slope(m) and the y-intercept(b).

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

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

Line BD:
[tex]m = \frac{-3-3}{-3-3}[/tex]
[tex]m = \frac{-6}{-6} [/tex]
∴ m = 1
∴ [tex] y_{BD} = 1x +b[/tex]

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:
[tex] y_{AC} = -x [/tex]
[tex] y_{BD} = x[/tex]

Hope this helps!