Question

i have two questions please


2)
Select the equation of a line that is perpendicular to the line on the graph and passes through the point (3, 2).

y = 1 over 3 x + 3

y = -1 over 3x + 3

y = 3 x + 2

y = - 1 over 3 x + 2

3)
Write the equation of the line that is parallel to the line y = -3x + 12 and passes through the point (-1, 6).

y = one thirdx + 7

y = -3x + 3

y = one thirdx + 3

y = -3x + 7

Answer 1

the graph is for the first one really need help.

Answer 2

Desmos graphing will help you with this :D

Answer 3

anyone

Answer 4

You need to find the slope of the line in the graph first.
Slope = rise / run. You need to pick two suitable points on the line to find the slope.

Answer 5

My estimate is the line crosses the x-axis at (-2/3, 0) and the y-axis at (0,2)
Slope = rise / run = y-intercept / x-intercept = 2 / (-2/3) = - 2 * 3/2 = -3

Slope of the perpendicular line will be the negative reciprocal of the slope.
Therefore, slope m = -1/3

y = mx + b
y = -1/3x + b

It passes through (3,2)
put x = 3, y = 2 and solve for b.

Answer 6

like i know it would be over something bc C a stupid answer

Answer 7

so the b would be the 2 right

Answer 8

y = -1/3x + b
It passes through (3,2)
when x = 3, y = 2
2 = -1/3 * 3 + b
2 = -1 + b
b = 3

y = -1/3x + 3

Answer 9

For #3) What is the slope of the line y = -3x + 12 ?

Answer 10

explain a little

Answer 11

y = -3x + 12
Compare it to the general formula y = mx + b where m is the slope.
Therefore, slope of the line y = -3x + 12 is ?

Answer 12

it would be y=-3x+7

Answer 13

nvm i got it