Find Equation, Graph, etc, given points (-2,-2), (5,-2)

[Jonathan Bower]

Given two Pts. (-2,-2) (5,-2)
Find Equation for the line?
Graph the line?
What is value of the slope?
Slope increase or Decrease?
Value of Y intercept?
I have tried everything and I can not figure it out. I believe it is a horiz. line. but the rest I can not figure out.

 

[stapel]

I have tried everything and I can not figure it out.

Please reply showing what you have tried. You plugged the two points into the slope formula, and... then what? You plotted the two points, and... then what? You plugged one of the points (which?) and the slope you found into a straight-line equation (which one?), and... then what?

Please be complete. Thank you!

Eliz.

 

[Jonathan Bower]

Thanks Eliz, I will show my work which has taken many routes
Given the two points (-2,-2) (5,-2)
Slope is change in Y over change in X so,
1. M=-2-(-2) /-2+-5 = -2+2/-7, =0/-7 M= - 0/7 slope is zero?

Y=Mx+b
y=0x+b
-2=5+b
-2-5=b
-7=b

I keep getting lost in what I come up with.

 

[stapel]

...slope is zero?

Yes; m = 0, as you'd expected when you plotted the two points.

Y=Mx+b
y=0x+b
-2=5+b <= How are you getting that (0)(5) = 5?
-2-5=b
-7=b

I keep getting lost in what I come up with.

I'm not sure why you're "lost"...?

You are using the slope-intercept form, y = mx + b, and you have found a value for m and, after making the correction above, a value for b; namely, m = 0 and b = -2. So you have the line equation as being y = 0 - 2, which obviously simplifies as y = -2. You've plotted the two points, and the value of the slope confirms your graph; this is indeed an horizontal line with zero slope. So you've found the slope, confirmed that it is neither positive nor negative, found the y-intercept, found the line equation, and done the graph.

Where are you stuck? Where did you get "lost"? Please be specific. Thank you!

Eliz.

Edit: Hadn't checked poster's math.

 

[Jonathan Bower]

I keep gettin diff. looking graphs. I will plot some x points again, fiving my some y points. I will graph it and attempt to explain my quandry. Math is not my best subject, at this point I do not know what my best subject is!
BRB. :shock:

 

[Jonathan Bower]

If I use
y-y1=m(x-x1)
y-(-2)=0(x-(-2)
y+2=0
y+2-2=0+(-2)
y=-2
Now I show y=-2

 

[stapel]

I keep gettin diff. looking graphs.

How many different lines are you getting through the two given points? :shock:

Now I show y=-2

Yes; this time, you got zero when you multiplied 5 by zero. :wink:

Eliz.

 

[Jonathan Bower]

I give up I quit.

 

[Deleted member 4993]

So equation of your line is:

y = -2.....................that's it!!

 

[Mrspi]

I give up I quit.

Don't quit....

Your two given points were (-2, -2) and (5, -2)

Plot those two points....then draw the line which contains them (as you guessed at first, it is a horizontal line). Any two distinct points determine exactly one line, so you can't be getting different lines when you draw the graph.

Now, you should have learned that the slope of ANY horizontal line is 0. And you should have learned that the equation of ANY horizontal line is of the form y = b, where "b" is the common y-coordinate for every point on the line.

In your case, every point has a y-coordinate of -2, so the equation is
y = -2