Question

What is the equation of a line that passes through points (-3,0.5) and (3, -0.5)?

A.) y = -1/6x
B.) y = -6x
C.) y = -1/6x + 1
D.) y = -6x - 17.5

Answer 1

y=-1/6x + 0 or y=-1/6x (when b = 0 then you leave it out)

Explanation:

Find the slope:

-0.5 - 0.5/3 - (-3) = -1/6

Find the y intercept (b): Plug in the x, y and m into y= mx + b

0.5=(-3)(-1/6) + b

0.5=1/2 + b

0.5-0.5 = b (change the 1/2 into a decimal 0.5 to make it easier to subtract

0=b

Now plug in the m and the b that you just found into the equation y=mx+b

y=-1/6x + 0 or y=-1/6x (when b = 0 then you leave it out)

Answer 2

The slope-intercept form:

`y=mx+b`

m - slope

b - y-intercept

The formula of slope:

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

We have the points (-3, 0.5) and (3, -0.5). Substitute:

`m=(-0.5-0.5)/(3-(-3))=(-1)/(6)=-(1)/(6)`

Put it to the equation of a line:

`y=-(1)/(6)x+b`

Put the coordinates of the point (3, -0.5) to the equation of a line:

`-0.5=-(1)/(6)(3)+b`

`-0.5=-(1)/(2)+b`

`-0.5=-0.5+b` add 0.5 to both sides

`b=0`

Answer:
`\boxed{A.)\ y=-(1)/(6)x}`