Question

Graph the line passing through (6, 1) whose slope is m = 3.

Answer 1

Answer: $y=3x-17$

Step-by-step explanation:

use point-slope form:

$y_{1} -y=m(x_{1} -x)$

$y-1=3(x-6)$

$y-1=3x-18$

$y=3x-17$

Answer 2

Answer:

equation: y = 3x - 17

[image attached]

Step-by-step explanation:

we know that the equation for a line is y = mx + b

we are given a point, (6, 1)

remember that a point is (x, y)

we can plug these sample x and y values into our equation, replacing x and y

 y = mx + b

 1 = m6 + b

we know that m = 3 [this was a given], so we can plug this into our equation

1 = (3)(6) + b

1 = (18) + b

1 = 18 + b       [we can now solve for "b"]

1 = 18 + b

-18    - 18

-17 = b

so, we can re-write our equation so that we can graph it:

y = 3x - 17

we can now plug in x-values to graph this:

(you'll see these written in the table on the graph)

y = 3(0) - 17

y = -17

y= 3(1) - 17

y = 3 - 17

y = -14

y = 3(2) - 17

y = 6 - 17

y = -11

y = 3(3) - 17

y = 9 - 17

y = -8

y = 3(4) - 17

y = 12 - 17

y = -5

y = 3(-1) - 17

y = -3 - 17

y = -20

y = 3(-2) - 17

y = -6 - 17

y = -23

y = 3(-3) - 17

y = -9 - 17

y = -26

[there is an image attached of my graph]

hope this helps!! :)