### Video Transcript

In this video, we’re going to look
at how to find the shortest distance from a point to a straight line in the
coordinate plane. So this is the situation that we’re
interested in. We have a coordinate plane and a
straight line and a point. And we’re looking to find the
shortest distance between the point and the line.

Now the first question might be,
“Well, how do I determine in what direction that distance should be?” There are lots of different
possible distances from the point to the line, depending which point on the line you
choose to connect to. A key fact that you need to be
aware of is that the shortest distance from a point to a line is the perpendicular
distance. So when we refer to the distance
between a point and a line, it’s this distance here that we’re looking to
calculate.

So I’ve chosen to label the
equation of the line in the form 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 is equal to 0. And I’ve chosen to label the point
they we’re interested in as having the coordinates 𝑥 one, 𝑦 one. There is a formula that we can use
in order to calculate the distance between the point and the line. And it’s this formula here. 𝑙, which represents the distance,
is equal to the absolute value or the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐, all
divided by the square root of 𝑎 squared plus 𝑏 squared.

Those modulus signs in the
numerator mean you take the absolute value of a quantity. So if it’s a positive value, then
it’s just that value. But if it’s a negative value, then
you multiply it by negative one. So, for example, the modulus of six
is six and the modulus of negative six is also six.

Now this formula is a handy
shortcut if you can remember it. But there is also a more complete
method that you may want to use, which is perhaps a little bit more intuitive. This method doesn’t just consider
that small distance 𝑙, but the full line which it’s part of. Now this line is perpendicular to
the line that you would be given in the question. So the first step is to find its
equation.

Remember, you know the coordinates
of one point on this line, 𝑥 one, 𝑦 one. You also know the slope of this
line because it’s perpendicular to the line 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero. You’d need to use the relationship
between the slopes of perpendicular lines, that is, that they multiply to negative
one. Once you’ve found the equation of
this line, you would then solve the equation simultaneously with the equation of the
first line in order to find the point of intersection of the two lines, which will
give the coordinates 𝑥 two, 𝑦 two.

Once you have the coordinates of
this point, we then use the distance formula in order to calculate the distance 𝑙,
the square root of 𝑥 two minus 𝑥 one all squared plus 𝑦 two minus 𝑦 one all
squared. That’s just an application of the
Pythagorean theorem. So as I mentioned, this method will
certainly be longer and perhaps more complex than using the formula that I mentioned
previously. But this is the method behind what
that formula was doing. So we’ll see both methods within
this video.

Find the length of the
perpendicular drawn from the point 𝐴 one, nine to the straight line negative
five 𝑥 plus 12𝑦 plus 13 equals zero.

So we’re going to answer this
question using the formula for calculating the distance between a point and a
straight line. So the formula is this. If I have the straight line
with equation 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 is equal to zero and I have a point with
coordinates 𝑥 one, 𝑦 one. Then the perpendicular distance
between them, 𝑙, is given by the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐, all
divided by the square root of 𝑎 squared plus 𝑏 squared. So what I need to do is
determine the values of 𝑎, 𝑏, 𝑐, 𝑥 one, and 𝑦 one and then substitute them
into the formula.

Let’s look at the straight line
first of all. I’m comparing it with 𝑎𝑥 plus
𝑏𝑦 plus 𝑐 is equal to zero. This shows me that 𝑎 is equal
to negative five, 𝑏 is equal to 12, and 𝑐 is equal to 13. Now let’s look at the point 𝐴,
which has coordinates one, nine. This tells me that 𝑥 one is
equal to one and 𝑦 one is equal to nine. So now I have all the values I
need. And it’s just a case of
substituting them into this formula for the distance 𝑙.

So we have that 𝑙 is equal to
negative five times one plus 12 times nine plus 13, the modulus of that
quantity. Then we’re going to divide it
by the square root of negative five squared plus 12 squared. This gives us the modulus of
negative five plus 108 plus 13 all divided by the square root of 25 plus
144. This gives the modulus of 116
over the square root of 169. Now as 116 is positive, then
its modulus is just its own value. So the numerator will be
116. And in the denominator, the
square root of 169 is 13 exactly.

So we have our answer to the
problem. The length of the perpendicular
between the point one, nine and the straight line negative five 𝑥 plus 12𝑦
plus 13 equals 0 is 116 over 13.

Find the length of the
perpendicular drawn from the point 𝐴 negative one, negative seven to the
straight line passing through the points 𝐵 six, negative four and 𝐶 nine,
negative five.

So the first thing we need is
we need to know the equation of the line that joins the points 𝐵 and 𝐶. So I’ll find this equation
using the point slope method. 𝑦 minus 𝑦 one equals 𝑚 𝑥
minus 𝑥 one. First, we’ll find the value of
𝑚, the slope of the line, using 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥
one. It doesn’t matter which way
round I allocate 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, so I’ve chosen to allocate
it in this order. It gives me negative five minus
negative four over nine minus six. This gives me a slope of
negative a third for this line.

Next, I’m gonna substitute one
of these points into the equation of the line. And again, it doesn’t matter
which I choose. I’ve chosen to substitute the
coordinates of 𝐵. So I’ve substituted six for 𝑥
one and negative four for 𝑦 one. This simplifies to give 𝑦 plus
four is equal to negative a third 𝑥 plus two. And finally, if I multiply the
whole equation through by three and group all the terms together on the
left-hand side, then it gives me the equation of the line, which is 𝑥 plus
three 𝑦 plus six is equal to zero.

So we could answer this problem
using the formula that’s on the screen, which tells us that the shortest
distance or the perpendicular distance between the line 𝑎𝑥 plus 𝑏𝑦 plus 𝑐
equals zero and the point 𝑥 one, 𝑦 one is given by the modulus of 𝑎𝑥 one
plus 𝑏𝑦 one plus 𝑐 all over the square root of 𝑎 squared plus 𝑏
squared.

However, I’m going to work
through the maths and the logic behind that formula in this question. So I’ve just drawn a sketch to
help visualise the situation. I’ve got the line 𝑥 plus three
𝑦 plus six equals zero and the point 𝐴 with coordinates negative one, negative
seven. And it’s this distance 𝑙 that
I’m looking to find.

The first thing I’m going to do
is find the equation of the line that connects 𝐴 to the line 𝑥 plus three 𝑦
plus six equals zero. There are now two things about
this line. I know the coordinates of one
point on the line, negative one, negative seven. And I know it’s perpendicular
to the line 𝑥 plus three 𝑦 plus six equals zero. So this is enough information
to enable me to find its equation.

As it’s perpendicular to the
first line, I can use the fact that the slopes of perpendicular lines are the
negative reciprocals of each other. They multiply to negative
one. Rearranging the equation of the
first line gives me 𝑦 equals negative a third 𝑥 minus two. So I can see that the slope of
the first line is negative a third. This means that the slope of a
perpendicular line must be three. So I’m going to find the
equation of this line using the point slope form. 𝑦 minus 𝑦 one equals 𝑚 𝑥
minus 𝑥 one.

I now know that 𝑚 is
three. And I know that the point I’m
going to use, the 𝑥 one, 𝑦 one, is the point with coordinates negative one,
negative seven. So this gives me 𝑦 minus
negative seven is equal to three 𝑥 minus negative one. This can be simplified in a
couple of lines of algebra to give the equation of the line 𝑦 equals three 𝑥
minus four.

So now I have the equations of
both lines. There are two more stages in
the method that I’m using here. The next stage is to find the
coordinates of the point of intersection of the two lines, the point that I’ve
labelled in orange on the diagram. To do this, I need to solve the
equation of the two lines simultaneously. I’m going to do this using
substitution, by substituting the expression for 𝑦 from equation two into
equation one. So this gives me 𝑥 plus three
lots of three 𝑥 minus four plus six is equal to zero.

Now there are a couple of lines
of algebra to work through in order to solve that equation for 𝑥. And I’ll leave you to fill
those in yourself. But if you do them correctly,
it will lead to 𝑥 is equal to three-fifths. Now we need to find the value
of 𝑦. So I’m gonna substitute 𝑥 is
equal to three-fifths into equation two. So we have 𝑦 is equal to three
multiplied by three-fifths minus four. And this gives us a value of
negative 11 over five.

So now we know the coordinates
of the point of intersection of the two lines. And there’s only one step left
in our method. We need to find the distance
between point 𝐴 and this point of intersection that we’ve just calculated. So in order to do this, we can
use the distance formula, which tells us that the distance 𝑙 is equal to the
square root of 𝑥 two minus 𝑥 one all squared plus 𝑦 two minus 𝑦 one all
squared. Where 𝑥 one, 𝑦 one and 𝑥
two, 𝑦 two represent the coordinates of the two points that we’re looking to
find the distance between. So substituting the values for
𝑥 one, 𝑥 two, 𝑦 one, and 𝑦 two gives me this calculation here for 𝑙.

Now if you evaluate this on a
calculator or if you work through the arithmetic yourself, it leads you to the
square root of 640 over 25. But this can be simplified. So we have a final simplified
answer of eight root 10 over five length units. You can confirm that this is
the same answer we would’ve achieved if we had used that standard formula
earlier within the question.

If the length of the
perpendicular drawn from the point negative five 𝑦 to the straight line
negative 15𝑥 plus eight 𝑦 minus five equals zero is 10 length units, find all
the possible values of 𝑦.

So this question is asking us
about the length of the perpendicular from a point to a straight line. So we need to recall the
standard formula for calculating this. The formula is this, that the
length of the perpendicular from the point 𝑥 one, 𝑦 one to the line 𝑎𝑥 plus
𝑏𝑦 plus 𝑐 equals zero is given by the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus
𝑐, all divided by the square root of 𝑎 squared plus 𝑏 squared.

We’re also told that, within
this particular question, this length is 10 length units. So let’s compare the general
information with the specific values in this question. If I look at the equation of
the line first of all, this tells me that 𝑎 is equal to negative 15, 𝑏 is
equal to eight, and 𝑐 is equal to negative five. Now if I look at the
coordinates of the points that we’re interested in, this tells me that 𝑥 one is
equal to negative five and 𝑦 one has just been given the general coordinate
𝑦.

I’m looking to find the
possible values of 𝑦. So I need to use this
information to set up and then solve an equation. So substituting those five
values into the relevant places in the formula and remembering that I’m told
that it’s equal to 10 gives me this equation here. Evaluating the numeric parts of
this tells me that I have the modulus of 70 plus eight 𝑦 divided by 17 is equal
to 10. And multiplying by 17, I have
that the modulus of 70 plus eight 𝑦 is equal to 170.

Now let’s look at what this
modulus means. It means the absolute value of
70 plus eight 𝑦. So 70 plus eight 𝑦 is either
equal to 170 or it could be equal to negative 170. This means that I have two
linear equations to solve, leading to two possible values of 𝑦. The first equation gives eight
𝑦 is equal to 100, and then 𝑦 equals 100 over eight, which simplifies to 25
over two. The second equation gives eight
𝑦 equals negative 240. And then dividing by eight
gives 𝑦 equals negative 30. So in this question, there are
two possible values of 𝑦. 𝑦 is either equal to 25 over
two or it’s equal to negative 30.

In summary then, we’ve seen that
the shortest distance between a point and a line is the perpendicular distance. We’ve seen the formula for
calculating this distance. 𝑙 is equal to the modulus of 𝑎𝑥
one plus 𝑏𝑦 one plus 𝑐, all divided by the square root of 𝑎 squared plus 𝑏
squared. We’ve also seen how to take a more
first-principles-like approach to answering such a problem by solving a system of
linear equations.