All of the points in this graph -- (1, 5), (a, 11), (6, b), (8, 19) -- are on the same line

[lu346]

All of the points in this graph are on the same line

 

[stapel]

All of the points in this graph are on the same lineView attachment 36618

How did you get 19 - 5 being equal to 18?

When you reply, please include a clear listing of your thoughts and efforts so far. (Read Before Posting)

Thank you!

 

[lu346]

I wasn’t really sure how to go about solving this. I tried to calculate slope but I think I did it wrong what would my first step be?

 

[stapel]

I wasn’t really sure how to go about solving this. I tried to calculate slope but I think I did it wrong what would my first step be?

Try plugging the two given points into the slope formula.

When you reply, please *show* your work, even if you think it is wrong. Thank you.

 

[Steven G]

I wasn’t really sure how to go about solving this. I tried to calculate slope but I think I did it wrong what would my first step be?

You know the formula for the slope as you tried using it. The only problem is that you made a slight mistake using it. Just try again and show us your work.

 

[lu346]

Rise=14 Run=7 so rise/run=14/7
Is 14/7 the slope?

 

[stapel]

Rise=14 Run=7 so rise/run=14/7
Is 14/7 the slope?

Yes, but surely this can be simplified?

 

[topsquark]

Rise=14 Run=7 so rise/run=14/7
Is 14/7 the slope?

Yes, but that could simply mean that pairs of points all lie on parallel lines. Have you come across an equation for a line in the xy plane? eg. y = mx + b?

-Dan

 

[pka]

All of the points in this graph are on the same lineView attachment 36618

You appear to be in a real confused mess. Two points determine an unique line.
The points $(8,19)~\&~(1,5)$ determine a line of slope $\dfrac{19-5}{8-1}=\dfrac{14}{7}=2$.
So the equation of the line is $\large\bf\boxed{2x-y+3=0}$.
Now you do some work and post answers.

$$$$

 

[lu346]

Here's what I did:
14/7=2 so the slope is 2
to find A: 11-5=6 and since the slope is 2/1, I did 6*1/2=3 so a=3
to find B: 8-6=2 and using the slope of 2/1, 2*2/1=4 so b=4
is this correct?

 

[pka]

Here's what I did:
14/7=2 so the slope is 2
to find A: 11-5=6 and since the slope is 2/1, I did 6*1/2=3 so a=3
to find B: 8-6=2 and using the slope of 2/1, 2*2/1=4 so b=4
is this correct?

To lu346. you seem to be lost in this question.
We are given that points $(1,5),~(a,11),~(6,b)~\&,~(8,19)$ are on the same line.
The slope of which is $m=\dfrac{19-5}{8-1}=\dfrac{14}{7}=2$ using the first & last given points.
The equation of the required line is $(y-5)=2(x-1)\text{ or }\large{\bf{\boxed{2x-y+3=0}}}$
Now you need to convince yourself that is indeed the equation of the line,
Using that equation we get $\left\{ \begin{gathered} 2 \cdot a - 11 + 3 = 0 \\ 2 \cdot 6 - b + 3 = 0 \\ \end{gathered} \right.$
So use those two to solve for $a~\&~b$.

 

[Harry_the_cat]

Here's what I did:
14/7=2 so the slope is 2
to find A: 11-5=6 and since the slope is 2/1, I did 6*1/2=3 so a=3
to find B: 8-6=2 and using the slope of 2/1, 2*2/1=4 so b=4
is this correct?

Yes the slope between the first and fourth point is 2.

So the slope between the first and second point must also be 2. ie$\displaystyle \frac{11-5}{a-1} =2$. Can you solve for $\displaystyle a$?

The slope between the first and third point must also be 2. Can you form an expression for the slope and then solve for $\displaystyle b$?

 

[lu346]

Yes the slope between the first and fourth point is 2.

So the slope between the first and second point must also be 2. ie$\displaystyle \frac{11-5}{a-1} =2$. Can you solve for $\displaystyle a$?

The slope between the first and third point must also be 2. Can you form an expression for the slope and then solve for $\displaystyle b$?

So 11-5=6 and 6*1/2=3 and since the point before it is out 1, that’s means that a=4
For b: 6-1=5 so run=5 and 5*2=10 so rise=10 so b=10

 

[blamocur]

For b: 6-1=5 so run=5 and 5*2=10 so rise=10 so b=10

Rise of 10 from which point?

 

[Harry_the_cat]

"So 11-5=6 and 6*1/2=3 and since the point before it is out 1, that’s means that a=4".

Correct.
In other words, solving the equation:
$\displaystyle \frac{11-5}{a-1}=2$

$\displaystyle \frac{6}{a-1}=2$

$\displaystyle 6=2\times (a-1)$

$\displaystyle \frac{6}{2} = a-1$

$\displaystyle 3=a-1$

$\displaystyle a=4$

Your answer b=10 is incorrect. Can you follow the example above to find b correctly? Be careful, it is a little bit different.
(Note that b could not possibly be 10 - look at the diagram - it must be more than 11 for starters. Always check the reasonableness of your answer.)

 

[pka]

Correct.
In other words, solving the equation:
$\displaystyle \frac{11-5}{a-1}=2$
$\displaystyle \frac{6}{a-1}=2$
$\displaystyle 6=2\times (a-1)$
$\displaystyle \frac{6}{2} = a-1$
$\displaystyle 3=a-1$
$\displaystyle a=4$

Surely it is simpler to see that $2a - 11 + 3 = 0 \Rightarrow \;a = 4~?$
AND that $2\cdot 6 - b + 3 = 0 \Rightarrow \;b =~?$
$$$$

 

[Dr.Peterson]

Surely it is simpler to see that $2a - 11 + 3 = 0 \Rightarrow \;a = 4~?$
AND that $2\cdot 6 - b + 3 = 0 \Rightarrow \;b =~?$

It might be, if you've already found the equation (and wrote it in your form, which is not what most beginners are taught, in my experience). This student may not have learned equations for lines yet at all. Since their focus is clearly on slopes, using that concept directly seems more likely to be helpful.

But that's why it's good when different people give different kinds of answers. We seldom know enough of the student's background to be sure what will help.

I myself might have done something more or less visual like this:

 

[Harry_the_cat]

Surely it is simpler to see that $2a - 11 + 3 = 0 \Rightarrow \;a = 4~?$
AND that $2\cdot 6 - b + 3 = 0 \Rightarrow \;b =~?$
$$$$

Why would you go to the bother of finding the equation of the line first? This is a harder concept that just working with the constant gradient.

 

[lu346]

Thank you, everyone, for you suggestions.

My teacher said we would have to solve it using slope, because we haven’t learned to use equations. So here is my final answer.

I am asked to find a & b.

I am given two points on the line: (1, 5) & (8, 19).

The slope of the line is, therefore, (19-5)/(8-1) = 14/7 = 2/1 = 2.

The slope tells me I must go 2 up for every 1 over (or 2 down for every 1 back).

Starting at (1, 5), to get to (a, 11) I must go 6 up (11-5), therefore, I must also go 3 over. So the x-coordinate of (a, 11) must be the x-coordinate of (1, 5) plus 3, therefore, (a, 11) = (1+3, 11) = (4, 11). So a = 4.

Starting at (1, 5), to get to (6, b) I must go 5 over (6-1), therefore, I must also go 10 up. So the y-coordinate of (6, b) must be the y-coordinate of (1, 5) plus 10, therefore, (6, b) = (6, 5+10)= (6, 15). So b = 15.

 

[blamocur]

Looks like you've nailed this one.