Question

Indicate the equation of the given line in standard form, writing the answer in the equation box below. The line containing the diagonal of a square whose vertices are AC(-3, 3), B(3, 3), C(3, -3), and D(-3, -3). Find two equations, one for each diagonal.

Answer 1

The equations of the diagonals of a square with given vertices are found by calculating the slope with opposite points and solving for the y-intercept using one of the points. The first diagonal has the standard form equation x + y = 0, and the second diagonal has the equation -x + y = 0.

Step-by-step explanation:

Finding the Equations for Diagonals of a Square

The diagonals of a square can be found by using the coordinates of opposite vertices. Given the vertices AC(-3, 3) and C(3, -3) for one diagonal, we can find the slope (m) of the diagonal by using the formula:

m = (y2 - y1) / (x2 - x1)

Substituting the coordinates, we get:

m = (-3 - 3) / (3 - (-3))

m = -6 / 6 = -1

The equation of a line in slope-intercept form is y = mx + b. We have m and can use either point to solve for b. Let's use point A(-3, 3):

3 = (-1)(-3) + b => b = 0

Hence, the equation of the diagonal AC in slope-intercept form is:

y = -x

To convert it to standard form, Ax + By = C, we can rewrite:

x + y = 0

Similarly, for the other diagonal BD, with vertices B(3, 3) and D(-3, -3), the slope is:

m = (3 - (-3)) / (3 - (-3))

m = 6 / 6 = 1

Using point B, to find b:

3 = (1)(3) + b => b = 0

The equation in slope-intercept form:

y = x

And the standard form:

-x + y = 0

We have found the equations of both diagonals in standard form, x + y = 0 and -x + y = 0.