Find the equation of the line that passes through points (-3,-1) and (2,1/2)

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (-3,-1), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=-3 and y1=-1.

Also, let's call the second point you gave, (2,1/2), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=2 and y2=1/2.

Now, just plug the numbers into the formula for m above, like this:

[tex]m=\frac{1/2--1}{2--3}[/tex] or

[tex]m=\frac{3/2}{5}[/tex] or

[tex]y={3/10x+b}[/tex]

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,-1). When x of the line is -3, y of the line must be -1.

(2,1/2). When x of the line is 2, y of the line must be 1/2.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=3/10x+b. b is what we want, the 3/10 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,-1) and (2,1/2).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want..the answer will be the same:

(-3,-1). y=mx+b or -1=3/10 × -3+b, or solving for b: b=-1-(3/10)(-3). b=-1/10.

(2,1/2). y=mx+b or 1/2=3/10 × 2+b, or solving for b: b=1/2-(3/10)(2). b=-1/10.

The euation of the line that passes through the points (-3,-1) and (2,1/2) is...