
Michael B. answered 11/16/12
I can provide your 'A-HA' moment
Hi Amber,
There are two main forms that you can write a line in:
Point-Slope: y-y1 = m(x-x1)
We use point-slope when we know the slope m and a point (x1, y1)
Slope-Intercept: y = mx + b
We use slope-intercept when we know the slope m and the y-intercept b
There is no fundamental difference between the lines produced by these two equations - in other words, we can convert between Slope-Intercept and Point-Slope form with a little bit of algebra.
Sometimes, however, the problems don't give us all the bits we need for either form directly, but we have to figure them out. That is the situation in this case, where we are given two points, but not the slope.
So the first step is to calculate the slope:
m = (y2 - y1) / (x2 - x1)
= (15 - -2) / (-3 - 14)
= 17 / -17
= -1
Now, since we know the slope, and we ALSO know two different points, we can plug one of the points and the slope into the point-slope formula:
y-y1 = m(x-x1) Starting equation
y-15 = -1(x- -3) Plug in m=-1 and (x1,y1) = (-3,15)
y-15 = -1(x+3) Minus a negative -> add a positive
But since the problem wants the answer in slope-intercept form, we must convert to this form using our basic algebra rules:
y-15 = -1(x + 3) Starting equation
y-15 = -x - 3 Distribute -1 across (x + 3)
y-15+15 = -x - 3 + 15 Add 15 to both sides
y = -x + 12 Combine terms, DONE
So the line, in slope-intercept form, is y = -x + 12.
Now you might ask why I plugged in (-3, 15) instead of (14, -2). Well it doesn't actually matter - we could have plugged in either point and we would have gotten the same answer in the end.