someone please help me i am so bad with point slope stuff
Write the equation of the line that passes through the points (2,−6) and (6,-6).
Write the equation of the line that passes through the points (9,8)and (6,3).
Write the equation of the line that passes through the points (-9,-4) and (-9,-4).
What is the equation of the line that passes through the point (−8,5) and has a slope of 0?
What is the slope of the line that passes through the points (-2, 7)and (2, -5).
What is the slope of the line that passes through the points (6, -5)and (6, -8).

Answer:

1.) y = -6

2.) y = 5/3x - 7

3.) y = 5

4.) y = -3x + 1

5.) x = -3

Step-by-step explanation:

So the equation you want to use is this:   [tex]\frac{y2-y1}{x2-x1}[/tex]

1.) Plug in

[tex]\frac{-6 - (-6)}{6 - 2} \\\\\frac{0}{4} = 0\\[/tex]

This line is zero slope to find the y-intercept get one of you points and plug them and the slope into this equation y = mx+ b. We will use (2,-6)

-6 = 0(2) + b (b represents y-intercept) (0*2=0)

-6 = b

So the equation is y = -6

2.) Do the same thing.

[tex]\frac{3-8}{6-9} \\\\\frac{-5}{-3} =\frac{5}{3}[/tex]

It's positive because when you divide a negative by a negative you get a positive. Plug in a point (6, 3) y = mx + b (b represents y-intercept)

3 = 5/3(6) + b (5/3*6 = 10)

3 = 10 + b (subtract 10 on both sides)

-7 = b

y = 5/3x - 7

3.) For this one you are just going to point slope form. y - y1 = m(x - x1)

Plug in

y - 5 = 0(x - (-8))

y - 5 = 0 (add 5 to both sides)

y = 5

4.) Do [tex]\frac{y2 - y1}{x2 - x1}[/tex] for this one

Plug in

[tex]\frac{-5 - 7}{2 - (-2)} \\\\\frac{-12}{4} = -3[/tex]

So the slope is -3. Plug into y = mx + b with the point (2, -5)

-5 = -3(2) + b (b represents y-intercept)

-5 = -6 + b (add 6 to both sides)

1 = b

The equation is y = -3x + 1

5.) Do the same thing as #4

[tex]\frac{-8 - (-5)}{6 - 6} \\\\\frac{-3}{0}[/tex]

This slope is undefined

Since you can't divide by 0 the equation would be

x = -3

Hope this helps ya!!

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