BAM!!! Mr. Tarrou. In this video we are going to do a couple of examples of finding the
distance between a point and a graph, or a function. Though a vertical line will pass
the function test, so I am going to say this is between a point and a line. The first example,
we are going to be able to look at and pretty much be able to tell what the answer is, because
I have given you a line which is vertical. The same would be true if the line given was
horizontal. You can really just look at the graph and tell what the distance between the
point and that line is. The second example is going to be dealing with an oblique line
which simply means that it has an angle. It is not vertical or horizontal. And we are
not going to be talking about this third scenario here. That would involving a lot of use Calculus.
But just to give you a purpose of learning this.... We are not going to be doing any
applications problems. This is just going to be algebraic examples. But this idea of
being given a graph, or in this case a parabola function and a point that is not on that function,
finding the distance between a point and a function... like here a point and a line,
is a concept that comes up repeatedly in your math studies. It will be coming up again.
You will be learning cool about derivatives which will tell you how to find the slope
of a curve. And you will find the slope of that curve which is the slope of tangent line.
You might have studied tangent lines. I am not sure. And we would be looking for this
distance between the point out in space and the point on on that function such that this
segment between the point on the function and this point out in space is got a perpendicular
slope to the tangent line. Now we are not studying Calculus in our Algebra video, but
this idea comes up over and over again. So there is a good reason to learn it. Let's
focus on this one first. We have a vertical line. Now vertical lines, well.. you know,
all graphs basically consist of an infinite number of points. And when you move just up
or down, it is only the y coordinate that is changing. So this is (2,0), (2,-1), (2,1),
(2,2), (2,3). So on a vertical line... the equation of a vertical line is just x equals
a number. So this the equation of x =2. Our point out in space is (5,2). Ok. Well I have
got a tape measure here, that is why it was in my hand. How do I want to measure the distance
between a point and a line? Or a parabola for that matter. Well, let's see. I don't
want to just run off the middle of nowhere. Think about your Mom or your Dad.. or somebody..
a sibling or someone taking care of you measures your height on the wall. They make a mark
and then they measure it perpendicular to the ground. Now this point is not 13 inches
away from this line... it is not 10 inches away from this line... it.. is.. well... We
are going to find the unit of measure in terms of my x and y axis. But at least the way I
have drawn it, it is about 7 and 1/2.. or 8 inches depending on where I put the beginning
of my tape measure on this dot. But, the point is you want to measure in a perpendicular
fashion. You want to find the closest distance, the true distance between this point and this
line. That is going to be measured in a perpendicular fashion. So if we have a vertical line of
x equals 2 and I want to measure the distance, then I am going to measure it horizontally.
Because, a horizontal line has a slope which is zero... or 0/1... and a vertical line has
a slope which is undefined... or just any about of rise over zero. The point is that
is perpendicular. Now horizontal and vertical distances, I can just count off. I am at the
x coordinate of 2... now I moved once and I am at the x coordinate of 3... I moved again
and I am at the x coordinate of 4... and finally I am at the x coordinate of 5. So this distance
between the point and the line is 5 (the x coordinate of this point) minus 2 which is
equal to 3. So! That is it! Right. That is a piece of cake, finding the distance between
a point and a line. Just look at the picture. I mean how hard could it be? Oh... at least
as far as linear functions we are about to find out. Here we have a line drawn about
as straight as I can get it. It is a little bit wiggly, or wobbly there. It has got the
equation of x minus 3y is equal to 6. And I have got a point in space which is (2,1).
I want to find out how far is that point from this line. Again we want to measure this in
a perpendicular fashion. So if I can guesstimate this as much as I possible can with a drawing...
I want to find the length of that orange line segment, the distance between... the perpendicular
distance... the shortest distance... between the point (2,1) and this line of x-3y=6. Well
this time I am not going to just be able to simply look at the answer... or look at the
drawing, excuse me, and come up with the answer. It is going to take some Algebra. It is going
to take some work. So what we are going to do is... We are going to find the equation
of this line. Now why do I want to find the equation of... well actually not this line...
This is only a segment. But I want to find the equation of the line that contains this
segment. Why do I need to do that? Because I want to find the distance between the point
and the line. Right. I just said that like three times. Well the distance formula is
equal to... the square root of x sub 2 minus x sub 1 squared plus y sub 2 minus y sub 1
squared. And the distance formula require you to know two points. And we are only given
one point. So we need to find the point of intersection between the line that contains
this orange segment and the line that is given. Find the intersection between two lines. That
is where that system of linear equations comes from that I was telling you about. So I need
to find the equation of a line which is perpendicular to the given line. So I need to look at that
3x-y=6 and manipulate it into a format that is easily... where I can look at it and easily
identify the slope. Now I also drew the picture, but let's just say that the picture was not
drawn and I cannot see the slope is equal to 1 over 3. I would take my equation, I will
put it into slope intercept form. That is how I like to do it anyway. So we are going
to subtract both sides by x and get negative 3y is equal to -x+6. Now I am going to divide
everything by -3. We get, let's just go this way.... Get y is equal to... negative one
divided by negative three is 1/3 x. Positive 6 divided by a negative 3 is -2. So I have
got my equation which was given to me in Standard Form into Slope Intercept Form. Now I can
see what the y intercept is... b... and I can see what the slope is. That is the coefficient
of x, the one-third. This is y=mx+b, the slope intercept form. I can look at this now and
see that my slope is equal to 1/3. Well, that is all fine and dandy, I do need that number,
and actually that number is the only thing really at this point... that I need from this
equation x-3y=6. Excuse the interruption there. So I want to find again the equation of the
line that contains that orange segment. Now that orange segment is perpendicular to the
given line. The given line has a slope of 1/3, so the perpendicular slope is the opposite
reciprocal of that white line. So I am going to take the 1/3 and I going to flip it and
make it 3 over 1 or just write 3. Because dividing by one does not do anything. This
has a positive and it is the opposite reciprocal. So it is -3. So there! I have got a enough
information. I have the minimum information that I need to write the equation of the line,
a point and a slope. I am going to use point-slope form which is Y-Y1 is equal to m times X-X1.
This is what I like to use to write the equation of a line. We have y minus... the given point
has a y value of 1, or y coordinate of 1. Our slope is negative 3. And then x minus
the x coordinate of our point which is equal to 2. We are going to distribute to get rid
of parenthesis. Y-1 is equal to -3x+6. Subtract 1 from both sides, or excuse me. It is already
subtraction, add one to both sides. And we get y is equal to -3x+7. This line contains
that orange segment. Now I still don't have my second point to find the distance with.
I just have the original (2,1). Now we need to find the coordinate of this intersection
point. We are going to do that by setting up a system of equations. The line containing
the orange segment and the original line which is again y is equal to 1/3 x - 2. Now I know
if you are watching this video and you are in this section of your studies that you have
learned how to solve linear systems with Substitution and linear systems with Linear Combination.
Honestly a lot of times I prefer the linear combination method where you manipulate your
equations so you have a set of opposite coefficients in front of one your variables. Then you add
them and make one of the variables disappear because when you add opposites you get zero.
Then you can continue on. But these equations are both solved for y. I don't care if they
are really both solved for y. I only need one of my equations solved for one of... or
a variable... one variable. And I am really kind of set up to use substitution right off
the bat. So we are going to solve this system of linear equations. I keep saying linear
because all of my exponents are one, and..hey... these are lines right. Linear functions. So
let's pick a blue chalk here. This equation is solved for y. So that 1/3 x - 2 which is
equal to y is going to get plugged into the other equation for y. You know, that is the
space where it gets put into. This is the substitution part of solving systems of linear
functions with substitution. We get one-third x minus 2... and any time you do substitution
you should wrap where that variable used to be with parenthesis and then plug in the expression
from the other equation. In this particular case there is nothing being multiplied to
the y. It is not like 2y. There is no division, there are no exponents. So the parenthesis
here are kind of overkill, but still not a bad idea or habit. So that 1/3 x - 2 is equal
to -3x+7. If have fractions in your equation and they make you uncomfortable and sometimes
you mess up a little with your fractions, just remember that a fraction bar is nothing
but division. So the inverse of division is multiplication. If you have a fraction in
your equation that make you uncomfortable, you don't like them, you don't want that division
in your equation, then do the opposite math operation and get rid of it. Let's go ahead
and multiply, since I already have the parenthesis here I am going to leave those there and add
a set over here... and I am going to multiply both sides of the equation by 3 to cancel
out that division of 3. Make it look like something that makes me feel more comfortable.
That is going to be equal to... 3 times 1/3 is... 3 times 1 is 3 over 3, which is 1. Or
the 3's cancel out giving me 1x. 3 times -2 is -6. 3 times -3 is -9. And then 3 times
7 is 21. Ok. I am trying to solve for x. I am trying to find the x coordinate of the
point where the two lines intersect. So I need to have an x on both sides... No, I need
to get them together. I am going to.. I will move the -9x to make things positive. We are
going to add 9x to both sides. That gives us 1 plus 9 is 10x... minus 6 just to show
all the steps. That equals 21. Undo... Ok, with this x now we are trying to solve for
we have a minus 6. I want to get rid of that negative 6 by doing the inverse math operation.
We are going to add both sides by 6. Undo that math operation of multiplication by dividing
both sides by 10 and get x is equal to 2.7. Now get to the point where I am going to have
to start referring to my notes a little bit to finish this problem up. Both of these line
segments are intersecting at the same point. Now I have the x coordinate of the point.
I need the y coordinate. It does matter because they are intersecting at the same point...
it does not matter if I put the 2.7 into the first equation.... either one of these linear
functions is fine. I am going to pick the one that at least does not have yet another
fraction in it. We are going to get y is equal to negative 3 times 2.7 plus 7. That is equal
to -3 times 2.7 plus 7 is equal to -1.1. So the intersection point of these two lines
is equal to (2.7,-1.1). Now before I erase the board and do the distance formula and
figure out the actual answer to this question, let me just... I tried to draw this somewhat
accurately. And are these points at least appearing to be reasonable from my sketch.
2.7 is somewhere between 2 and 3. It does look like my intersection point is somewhere
between 2 and 3. My y coordinate is a little bit less than -1. And actually it seems like
I am quite far below -1, but this was on... I did not use a ruler or graph paper. So at
least my intersection point is between -1 and -2. So my answer seems reasonable. Now
I finally have two points. The point of intersection, the given point, and I am out of space. Let
me get out of the way and this get this board erased. Alrighty then! [laughing] Point 1,
point 2. Let's do the distance is equal to the square root of... and you know, I said
point one and then point 2. Let's just make sure we keep that notation as clear as possible.
Not accidentally put a number in the wrong spot. And ok.. Well if I said one and two,
I am going to x sub 2 which is 2.7... minus x sub 1 which is equal to 2... I always us
subtraction when I am substituting anything like I said... Especially with subtraction
I mean I am going to use those parenthesis when I plug those numbers in to make sure
I don't make any sign errors. And we have y sub 2 which is -1.1... minus... I am not
using parenthesis, though it is really not a bad idea, simply because there is nothing
else out front. Like if I had labelled these points in reverse I would have had something
minus that negative 1.1. I would have to remember that the negative from the formula and the
negative from what I plugged in would cancel out. But, actually I really intended on showing
that... But I already made my labels so there you go. Y sub 2 minus Y sub 1... squared.
That is going to come out to be the square root... 2.7 minus 2 is equal to .7... plus
negative 1.1 minus 1 is -2.1. Now I have to revert back to my notes again. Positive .7
squared is .49. Actually... Yes, I don't really need those parenthesis. Well, I will keep
them for good measure.... Plus... Ok. Make sure that when you are squaring negative numbers
remember there is a difference between -2^2 and (-2)^2. Our two and three line scientific
calculators and our graphing calculators will use the notation you put into it correctly.
[phone ringing] And it will do that right after I get the phone. Alright. So negative
two squared is negative four. Whereas (-2) squared wrapped in the parenthesis is -2 times
-2 which is positive 4. Alright! This negative 2 squared without parenthesis, you are only
squaring the 2 and not the negative out front which is why the answer is negative 4. So
again careful use of your parenthesis. (-2.1) squared is positive 4.41. And let's get this
out of here. That comes out to be the square root of 4.9. And the square root of 4.9 is
approximately equal to 2.21 units. And that is the end of my example. And that is about
as hard as these problems will get. Unless of course you are getting into calculus where
you are finding say the distance between a point not on a parabola and the parabola itself.
So I am Mr. Tarrou. BAM! Go Do Your Homework!!! Oh, by the way if you want to see some of
those Calculus problems they are called Optimization or they are Optimization problems and I will
have a video on them coming up shortly:)