Question

Write an equation in slope-intercept form for the line with slope 3 that contains the point (5,-1).

Answer 1

The point-slope form:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - point

We have the slope m = 3 and the point (5, -1). Substitute:

$y-(-1)=3(x-5)\\\\y+1=3(x-5)$

Convert to the slope-intercept form: $y=mx-b$

$y+1=3(x-5)$          use distributive property a(b + c) = ab + ac

$y+1=3x-15$           subtract 1 from both sides

$\boxed{y=3x-16}$

Answer 2

Answer:

y = 3x-16

Step-by-step explanation:

We know the slope and a point , so we can use the point slope form of the equation for a line

y-y1 = m(x-x1) where the slope is m and the point is (x1,y1)

y--1 = 3(x-5)

y+1 = 3(x-5)

We need to rearrange the equation into slope intercept form  y=mx+b

First distribute the 3

y+1 =3x-15

Then subtract 1 from each side

y+1-1 =3x-15-1

y = 3x-16

This is in slope intercept form