Question

What is the standard form equation of the line shown below?

Graph of a line going through negative 2, 3 and 1, negative 3

a. y + 3 = −2(x − 1)
b. y = −2x − 1
c. 2x + y = −1
d. −2x − y = 1

Answer 1

The point-slope form:

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

m - slope

The formula of a slope:

`m=(y_2-y_1)/(x_2-x_1)`

We have the points (-2, 3) and (1, -3). Substitute:

`m=(-3-3)/(1-(-2))=(-6)/(3)=-2\\\\y-(-3)=-2(x-1)\\\\y+3=-2(x-1)`

The standard form: Ax + By = C. Convert:

`y+3=-2(x-1)` use distributive property

`y+3=-2x+2` subtract 3 from both sides

`y=-2x-1` add 2x to both sides

`\boxed{2x+y=-1}\to\boxed{c.}`

Answer 2

b

Explanation:

You can determine the gradient from the two points (-2, 3) and (1, -3). The gradient of a straight line is:

`m=(y_2-y_1)/(x_2-x_1)`

`m=(-3-3)/(1-(-2))=-6/3=-2`

The equation for a straight line is:

`y=m\cdot{x}+c`

c is the y-intercept when x=0. We can substitute any point into the equation. Lets use point 1 and solve for c .

`3=-2\cdot{-2}+c`

`c=-1`

The equation is:

`y=-2\cdot{x}-1`

The answer is b