Let's explore distances on a number line.
Find all points 3 units away from 7
on a number line. Let's draw it first.
Here's our number line. You can only go
in two directions so it looks like we're
either going to add 3 to 7 or
subtract 3 from 7, right? That
means the two points would be 10 and
4. So there's two solutions to this
question. Let's look at something
different. In general, there is a distance
formula for numbers on a number line.
It goes like this. If we have any two
points A and B the distance between them
is a positive difference between A and B.
So, in this situation where B is to the
right of A, we would do B minus A. But if
we don't really know the order of the
points, then we have to do B minus A and
also A minus B and ask ourselves which
one is positive, because distance is
positive. The other option is to use the
absolute value. If you use the absolute value of A minus B or B minus A you get a
positive result. So you have two choices:
you either work it out two different ways
or you can use the absolute value
symbols. If you work it out two ways, take
the positive answer. So let's use what we
know about distance on number lines to
find out how many points are the same
distance away from 3 as they are
from 10. So let's draw it.
Let's check each of the regions in the
number line. Can a point out here be the
same distance from 3 as it is from
10? No, that's not going to work it's
too far from 10.
The same is true over here right? It's going to be too close to 10 and too far from
3. The only way to find a point
that's the same distance from both is
the point exactly halfway: the midpoint.
To find the midpoint you can just find
the average of 3 and 10, add them,
and divide by 2.
There it is. The number halfway between
them is 7.5. That's the only number on
the number line that is equidistant from
both 3 and 10, so there's only one
solution to that problem.
This one's really interesting to think
about. How many points are twice as far
from 2 as they are from 5?
Well, let's consider the possible regions
on the number line. So here's 2 and
here's 5. Is it possible for a
solution to be over here?
Remember, we're looking for a point that is twice as far from 2 as it is from 5.
This one's closer to 2 than it is to 5, so that's not possible.
How about in between? Is it possible to
put a point somewhere in between so it's
twice as far from 2 as it is from 5?
About right about there. Yeah, that seems
plausible,
so let's try to solve for that. How about
beyond the 5? Is it possible to have a
number out here that's twice as far to 2
as it is to 5?
Yeah, that seems possible, so it looks
like there's probably two solutions.
Let's write some equations and see if we
can solve for each of these. Let's call
the first one A
and the second one B. For the first one,
remember the distance from 2 to A is
twice as far as the distance from A to 5.
So let's write an equation that says
that. The distance from 2 to A has to be
written A minus 2, because remember A is greater than 2 in this case. That
distance is (remember is and equals go
together) is 2 times the distance from
5 to A. 5 is bigger so I'm going to
write 5 minus A to get a positive result.
Okay, now we solve for A.
If we simplify the right side, we get
10 minus 2A. Okay well I'm going to add the 2A
over to the side and add the 2 also.
See what we've got now. 3A is equal to 12
Well that's easy to solve: so A must
be 4. Let's double check and see if
that makes sense
according to our diagram. Could this be
4 right here?
Sure, that makes sense it's between 2
and 5.
Okay, let's try to solve for B. So an
equation for B that still means the
distance from B to 2 (B is greater than 2
so i'm going to write B minus 2) is
twice as big as the distance from B to 5.
B is greater than 5 so I'm going to say
B minus 5 to get that positive result.
Let's simplify. Okay get those B's
together by subtracting 2B. It's totally
fine to move the B over to the 2B if
you like doing it that way. So I get
negative B equals negative 8, if I
divide both those by negative 1,
I get B equals 8. Let's see if that looks
right on the diagram
could this be 8 right here? 8 is
greater than 5 so now let's check the
distances. 8 is three units from 5 and
8 is six units from 2.
Yeah, that's twice as far and 4 is one
unit from the 5 and two units from
2, so that also works.
Don't forget to check your solutions
because just because you made it work
with an algebra equation is not proof
that you did it correctly, so it's always
important to check constantly to see if
our mathematics seems reasonable and if
we can plug it back in and see if it
works that's even better.
And that's kind of the beginning
of exploring distances on a number line
