Prove whether the quadrilateral is a parallelogram using the Distance & Slope Formulas D(-5, -6), E(5, 2), F(4, -4), G(-6, -12)

Respuesta :

Answer:

See the distances bellow

Step-by-step explanation:

Given data

D(-5, -6), E(5, 2)

substitute into the expression for the distance

[tex]d= \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)\\\\d= \sqrt((5+5)^2 + (2+6)^2)\\\\d= \sqrt((10)^2 + (8)^2)\\\\d= \sqrt(100 +64)\\\\d= \sqrt((164)\\\\d= 12.80[/tex]

E(5, 2), F(4, -4)

[tex]d= \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)\\\\d= \sqrt((4-5)^2 + (-4-2)^2)\\\\d= \sqrt((-1)^2 + (-6)^2)\\\\d= \sqrt(1+36)\\\\d= \sqrt((37)\\\\d= 6.08[/tex]

F(4, -4), G(-6, -12)

[tex]d= \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)\\\\d= \sqrt((-6-4)^2 + (-12+4)^2)\\\\d= \sqrt((-10)^2 + (-8)^2)\\\\d= \sqrt(100+64)\\\\d= \sqrt((164)\\\\d= 12.80[/tex]

G(-6, -12),  D(-5, -6)

[tex]d= \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)\\\\d= \sqrt((-6+5)^2 + (-12+6)^2)\\\\d= \sqrt((-1)^2 + (-6)^2)\\\\d= \sqrt(1+36)\\\\d= \sqrt((37)\\\\d= 6.08[/tex]