Question

I'm confused again...

Instead of the y = mx + b I'm given no formula just told to write the equation for the following relation.

which is: B = {(x,y): (2,3), (4,4), (6,5)}
The answer is 2y = x + 4

I tried the y = mx + b formula, which turn to be -

(4-3)/(4-2)= 1÷2=2
(5-4)/(6-4)= 1÷2=2
(5-3)/(6-2)= 2÷4=2

now its y = 2x + b

3=2(2) + b
-4 + 3 = b
-1 = b
It would all be y = 2x -1
but since i know the answer its not that, is it a different formula and if so how do you know when to use the right formula when they don't tell you which one?

Answer 1

I think you mean "write the formula for the following _relation_."

Answer 2

We can still use y = mx + b
First, we use any pair of 2 of the given points to find the slope.
Do you know how to find the slope of a line given two points on the line?

Answer 3

The slope, \(m\), of the line that passes through points \((x_1, y_1)\) and \((x_2, y_2)\) is
\(m = \dfrac{y_2 - y_1}{x _2 - x_1} \)

Answer 4

first find m= (y-y0)/(x-x0) choice any 2 points

Answer 5

then use y-yo=m(x-x0) use any points

Answer 6

i used 1 and 2 points y=1/2x+2

Answer 7

multply this by 2
2y=x+4

Answer 8

Once you find the slope, plug it in the equation
y = mx + b
Then use any point and enter its coordinates for x and y in the equation above (also using the value of m you obtained above), and solve for b. Then plug in b into the equation.

Answer 9

so its \[M = \frac{ 3-4 }{ 2-4 }= \frac{ 1 }{ 2 }\]

I'm stuck on y-yo=m(x-x0)

would it be written 3-4 = m(2-4) or do I add that 0 by x0 to my 4 making it 40?

but isn't the slope 1/2 so I'd go to that y = mx+b

I end up with 3 = 2(2) + b
-4+3=b
1/2-1=b
-0.5=b
which isn't right...

Answer 10

No.
The idea of y - y0 = m(x - x0)
is to leave x and y as they are, and use the point's coordinates in x0 and y0.
Like this:
y - 3 = (1/2)(x - 2)

Answer 11

The above was using the correct slope you found of 1/2 and the point (2, 3)

Answer 12

So now I'd bring in the y = mx + b?

Making it... y - 3 = (1/2)(2) + b
y -3=1+b
-1-3=b
2=b

then it'd be 2y = x... and where's the 4...

or is it y = 1/2x + 2(2)
y=1/2x + 4

then we do y-3=(1/2)(2) + 4
-1-3=4
2=4
then write it out like 2y = x + 4... and if so, do I leave x as it is, from the y - y0 = m(x - x0) formula since one x was left alone and the same with 2y since one y was left alone?

Answer 13

There are several ways of doing this problem.
I'll show you two ways.
1) Using \(y = mx + b\)
2) Using \(y - y_0 = m(x - x_0) \)

For both ways you first need to find m, the slope, so let's start with that.
I am using the first two given points, (2, 3) and (4, 4).:
\(m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{4 - 3}{4 - 2} = \dfrac{1}{2}\)
You already had the slope above that you calculated correctly.
Now let's follow method 1) y = mx + b
First, we plug in the slope into the equation y = mx + b to get
\(y = \dfrac{1}{2}x + b\)
We need to find b.
We use any point of the three given points in y = mx + b and solve for b.
Let's use point (2, 3)

\(3 = \dfrac{1}{2}\times 2 + b\)

\(3 = 1 + b\)

\(2 = b\)

Now that we have b = 2, we substitute this value into \(y = \dfrac{1}{2}x + b\) to get:

\(y = \dfrac{1}{2}x + 2\) This is the correct equation in slope-intercept form.

Now we multiply both sides by 2 to get:

\(2y = x + 4\) which is the answer you know is correct.

Answer 14

Now let's use method 2) \(y - y_0 = m(x - x_0)\)
Here we need the slope, m, which we already have, and we need a point. The coordinates of the point we choose to use are substituted into \(x_0\) and \(y_0\).

\(y - y_0 = m(x - x_0)\)

Let's use the last given point, (6, 5)

\(y - 5 = \dfrac{1}{2}(x - 6) \)

Multiply both sides by 2:

\(2y - 10 = x - 6\)

Add 10 to both sides:

\(2y = x + 4\)

Once again, as expected, we find the same answer.

Answer 15

This is how you solve this problem using these two methods.

Answer 16

In your last response, you were using y = mx + b.
You did calculate b correctly as b = 2.
Then you did not know how to apply the b = 2.
I hope the explanation above clears this up.
If you still have any questions, feel free to ask.

Answer 17

I think I have it, I'll test it out on a few other problems to make sure I got it, and thank you for the help.

Answer 18

You're welcome.