Answer:
Step-by-step explanation:
With any two points, you can find the line between them with:
[tex]y = mx + b\\where\\m = \frac{y_2-y_1}{x_2-x_1}[/tex]
#1:
(-1, 1) -> (1, 3) [tex]m = \frac{3-1}{1-(-1)} = 1, 3 = 1(1) + b, b = 2, y = x + 2[/tex]
(3, 4) -> (0, 2) [tex]m = \frac{4-2}{3-0} = \frac{2}{3}, 2 = \frac{2}{3}(0) + b, y = \frac{2}{3}(x)+2[/tex]
(0, 1) -> (3, 3) [tex]m = \frac{3-1}{3-0} = \frac{2}{3}, 1 = 2/3(0) + b, y = \frac{2}{3}(x) + 1[/tex]
Lines b and c are parallel because they have the same slopes
#2 is a bit easier
a: 2y = x + 12, [tex]y = \frac{1}{2}(x) + 6[/tex]
b:2y - x = 5, [tex]y = \frac{1}{2}(x) + \frac{5}{2}[/tex]
c: 2y + x = 4, [tex]y = \frac{-1}{2}(x) + 2[/tex]
Since lines a and b have the same slope, they are parallel
#3:
(1, 3), y = 2x - 5
We have to find a line with a slope of 2 that passes through the two points
[tex]y = 2x + b\\3 = 2(1) + b\\b = 1\\\\y = 2x + 1[/tex]
#4:
(-2, 1) , y = -4x + 3
We have to find a line with a slope of-4 that passes through the line
[tex]y = -4x + b\\1 = -4(-2) + b\\1 = 8+ b\\b = -7\\y = -4x - 7[/tex]
#5:
(-2, 3) (1, -1), [tex]m = \frac{3-(-1)}{-2-1} = \frac{-4}{3}, 3 = \frac{8}{3} + b, b = \frac{1}{3}, y = \frac{-4x}{3} + \frac{1}{3}[/tex]
(-3, 1) (1, 4), [tex]m = \frac{4-1}{1-(-3)} =\frac{3}{4}, 4 = \frac{3(1)}{4} + b, y = \frac{3}{4}(x) + \frac{13}{4}[/tex]
(0, 2) (3, -2) [tex]m = \frac{2-(-2)}{0-3} = \frac{-4}{3} , 2 = b, y = \frac{-4}{3}(x) + 2[/tex]
Since lines c and b have negative reciprocal slopes, they are perpendicular
#6:
a: y = -4x + 7
b: x = 4y + 2, [tex]y = \frac{x}{4} - \frac{1}{2}[/tex]
c: -4y + x = 3, [tex]y = \frac{x}{4} - \frac{3}{4}[/tex]
since lines a and c have negative reciprocal slopes, they are perpendicular
