Question

indicate the equation of the given line in standard form show all your work for full creditthe line containing the diagonal ,BD,of a square whose vertices areA(-3,3) B(3,3) C (3,-3) and D (-3,-3)find two equations one for each diagonal

Answer 1

ANSWER:

x + y = 0

x - y = 0

EXPLANATION:

Given the coordinates:

A(-3,3)

B(3,3)

C(3,-3) and

D (-3,-3)

One diagonal is line AC.

Find the slope of AC:

$m\text{ = }\frac{y2-y1}{x2-x1}=\text{ }\frac{-3-3}{3-(-3)}=\frac{-6}{6}=\text{ -1}$

Solpe of AC = -1

Using the slope intercept form, y = mx+b, we have:

y = -1x + b

y = -x + b

The equation for AC:

y = -x

The other diagonal is BD.

find the slope of BD:

$m\text{ = }\frac{y2-y1}{x2-x1}=\text{ }\frac{-3-3}{-3-3}=\frac{-6}{-6}=\text{ 1}$

Slope of BD = 1

The equation of BD:

y = 1x

y = x

Therefore the equation of both lines in standard form will be:

AC ==> x + y = 0

BD ==> x - y = 0