Answer:
Y= -1/2x+2
Step-by-step explanation:
First of all, remember what the equation of a line is:
y = mx+b
Where:
m is the slope, and
b is the y-intercept
First, let's find what m is, the slope of the line...
The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.
For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:
So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (0,2), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=0 and y1=2.
Also, let's call the second point you gave, (4,0), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=4 and y2=0.
Now, just plug the numbers into the formula for m above, like this:
m=
0 - 2
4 - 0
or...
m=
-2
4
or...
m=-1/2
So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:
y=-1/2x+b
Now, what about b, the y-intercept?
To find b, think about what your (x,y) points mean:
(0,2). When x of the line is 0, y of the line must be 2.
(4,0). When x of the line is 4, y of the line must be 0.
Because you said the line passes through each one of these two points, right?
Now, look at our line's equation so far: y=-1/2x+b. b is what we want, the -1/2 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 (0,2) and (4,0).
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:
(0,2). y=mx+b or 2=-1/2 × 0+b, or solving for b: b=2-(-1/2)(0). b=2.
(4,0). y=mx+b or 0=-1/2 × 4+b, or solving for b: b=0-(-1/2)(4). b=2.
See! In both cases we got the same value for b. And this completes our problem.
The equation of the line that passes through the points
(0,2) and (4,0)
is
y=-1/2x+2
on the matrix gives![\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%265%265%5C%5C1%26-%5Cdfrac23%265%5Cend%7Barray%7D%5Cright%5D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26-%5Cdfrac23%265%5C%5C5%265%265%5Cend%7Barray%7D%5Cright%5D)
