Video Transcript
In this video, weβll be looking at
and working with parallel and perpendicular lines. Weβll go through the definitions of
the terms parallel and perpendicular, and then weβll look at their equations and go
over some questions. Two lines are said to be parallel
if theyβre in the same plane, and no matter how far you extend them in each
direction, theyβre the same distance apart. Another popular definition also
says that they must never intersect. And the effect of that definition
is that if you have two lines that are in fact the same line, one on top of each
other, theyβre not parallel because they intersect in an infinite number of
different places. So although theyβre always the same
distance apart, zero, we donβt count them as being parallel because they do
intersect.
Now hereβs an example where weβve
got two different lines. Theyβre both on the π₯π¦-coordinate
plane and theyβre parallel. Wherever you measure the distance
between them, youβll always get the same answer. Two lines are said to be
perpendicular if a right angle is formed at their point of intersection, like it is
here. So these two lines are both on the
π₯π¦-coordinate plane and they intersect at the point two, zero. And the angle between them there is
90 degrees. In the rest of this video, weβre
gonna be working only with lines in the π₯π¦-coordinate plane. And weβll be looking at the
equations of these lines, picking out aspects that tell us if theyβre parallel or
theyβre perpendicular.
And equations of straight lines are
really an important part of this topic, so letβs just do a quick recap of π¦ equals
ππ₯ plus π. So the general form of an equation
of a straight line is π¦ equals ππ₯ plus π or maybe π¦ equals ππ₯ plus π,
depending on where you live. The π-value, the multiple of π₯,
the coefficient of π₯, tells you the slope of the line. And the value of π tells you where
it cuts the π¦-axis. Now the slope, or the gradient, of
the line is defined as being the amount that the π¦-coordinate changes by, when we
increase the π₯-coordinate by one.
Now, Iβve drawn two points π and
π on my line and Iβve increased the π₯-coordinate by one to get from π to π. So the difference in their
π₯-coordinates is positive one. And when I do that, the difference
in the π¦-coordinates from π to π will be an increase of π. So whatever value this is, okay,
that is the amount that the π¦-coordinate increases by, every time I increase my
π₯-coordinate up by one, moving along the line. So if that line had been for
example π¦ equals 0.5 π₯ plus three, the multiple of π₯, the slope, the π-value is
0.5. This means that every time I
increase my π₯-coordinate by one, my π¦-coordinate will increase by 0.5.
So if we arrange the equation of
the line in this format π¦ equals ππ₯ plus π, then itβs a really simple matter to
read off the slope of that line. Just look at the coefficient of π₯,
the multiple of π₯; that is your slope. So here we have a question.
Which of the following straight
lines have the same slope or gradient? And then weβve got five lines. A is π¦ equals three π₯ minus
seven. B is π¦ equals minus a half π₯ plus
three. C is π¦ equals minus three π₯ plus
seven. D is π¦ equals minus a half π₯ plus
five. And E is π¦ equals three π₯ plus
nine. Now, all of those equations are
already in the π¦ equals ππ₯ plus π format for us, so what we need to do is look
at the π₯-coefficients. And if theyβre the same number with
the same sign, then theyβll have the same slope or gradient.
So B and D have the same slope, or
gradient, of negative a half. And A and E have the same slope of
three. C has got a slope of negative
three, so itβs a different sign to A and E. So that is not the same slope or
gradient. Now when two lines have the same
slope or gradient, we call them parallel. So here, A is parallel to E and B
is parallel to D.
So now we can recognise parallel
lines just by looking at their equations, so long as theyβre in the π¦ equals ππ₯
plus π format. So thinking about the slope of
perpendicular lines, if two lines are perpendicular, then the product of their
slopes, or gradients, is negative one.
Well, why would that be? Letβs take a look. So Iβve got two lines here, πΏ one,
which Iβve given the equation π¦ equals ππ₯ plus π, and πΏ two, which Iβve given
the equation π¦ equals ππ₯ plus π. Letβs take that point of
intersection and then a point on each of those lines which has an π₯-coordinate
which is one greater than that π₯-coordinate at the point of intersection. For line one, the difference in the
π¦-coordinates between those two points, this one and this one, is going to be
π. And for line two, the difference in
those coordinates is going to be π. Well in fact, itβs going to be
negative π. So line two is a downhill line,
which means itβs going to have a negative gradient.
Now Iβm interested at the moment in
the distance here. And that is a positive
distance. So I need to take the negative of
that negative gradient to work out what that actual distance is going to be. Now if I label those points A, B,
and C, we know that triangle ABC is a right-angled triangle because the two lines
are perpendicular. So the angle ABC is a right
angle. And in right-angled triangles, we
can use the Pythagorean theorem to say that the square of the hypotenuse is equal to
the sum of the squares of the other sides. So length AC squared is equal to
length AB squared plus length BC squared. And we know that length AC is π
plus negative π. So AC squared is π plus negative
π all squared. Now if we look back at our diagram,
we can see that weβve got two more right-angled triangles. Weβve got this one here and this
one here.
And again, we can use the
Pythagorean theorem to work out the length of the hypotenuse of each of those. We know that the top triangle has a
height of β and a width of- sorry, a height of π and a width of one. And the bottom triangle has a
height of negative π and a width of one. So those lengths AB and BC are the
square root of π squared plus one squared and the square root of negative π all
squared plus one squared. And now we can start to simplify
this. π plus negative π is just π take
away π. So the left-hand side becomes π
minus π all squared. The square root of π squared plus
one squared all squared is just π squared plus one. And negative π all squared is just
π squared. So the square root of negative π
all squared plus one squared all squared is just π squared plus one.
Now π minus π all squared means
π minus π times π minus π and just tidying up the right-hand side, Iβve got π
squared plus π squared plus two. Now multiplying each term in the
first bracket by each term in the second bracket on the left-hand side, I get π
squared minus two ππ plus π squared. So I can subtract π squared from
both sides and I can subtract π squared from both sides, which eliminates π
squared from both sides and eliminates π squared from both sides. So Iβve got minus two ππ is equal
to two. Now if I divide both sides by
negative two, I get π times π is equal to two over minus two, which is equal to
negative one. Now remember, π was the slope of
my first line and π was the slope of my second line. So ππ is the product of the two
slopes of the lines. So when those lines were
perpendicular, it doesnβt matter what the actual slopes were, we knew that when we
multiply them together, weβll always get this answer of negative one.
Now if you got a bit lost along the
way during that explanation, donβt worry. That bitβs not important. This is what youβve got to
remember. If two lines are perpendicular,
then the product of their slopes, or gradients, is negative one. So if we call π the slope of the
first line and π the slope of the second line, π times π is equal to negative
one. Or if I divide both sides by π, I
get π is equal to minus one over π. Or if I divide both sides by π, I
get π is equal to minus one over π. In other words, each slope is the
negative reciprocal of the other. This means that if I know one of
the slopes, I can find the perpendicular slope by just changing the sign and
flipping that number.
So for example, if π was five, π
would just be negative one over five, the opposite sign and then flip that
number. If π was minus three, I take the
opposite sign to make it positive and flip that number, three over one becomes one
over three. And if π was equal to two-thirds,
π will be negative three over two. So knowing this rule means that if
you know the gradient, or the slope, of a particular line, itβs very easy to work
out what the gradient, or the slope, of a perpendicular line to that one would
be.
So to sum up the basic facts, two
lines are parallel if they have the same slope but different π¦-intercept. For example, π¦ equals seven π₯
minus five and π¦ equals seven π₯ plus two. They both have a slope of seven and
their π¦-intercepts are different, minus five and positive two, so theyβre
parallel. And two lines are perpendicular if
the product of their slopes is equal to one. For example, π¦ equals three π₯
minus one and π¦ equals minus a third π₯ plus nine. The slope of the first is three and
the slope of the second is negative a third. So the product of those slopes is
three times negative a third. But three is the same as three over
one, so to make this a fraction calculation, Iβve got three times negative one over
one times three. Well thatβs negative three over
three, which is equal to negative one. So it meets the criteria, so the
lines are perpendicular.
And one more example, π¦ equals
negative two-sevenths π₯ plus eight and π¦ equals seven over two π₯ plus eight. The slopes are negative two over
seven and seven over two. Theyβre the negative reciprocal of
each other. And if I multiply the slopes
together, negative two over seven times seven over two is equal to negative fourteen
over fourteen which is equal to negative one. So those two lines are
perpendicular. And with perpendicular lines, it
doesnβt matter that those intercepts were both at positive eight. The slopes are different, so they
are different lines. They are definitely
perpendicular. Letβs have a look at some typical
questions then.
Two lines, A and B, have slopes
three-quarters and minus four over three, respectively. Are they parallel, perpendicular,
or neither?
Well, the slopes arenβt equal, so
theyβre definitely not parallel. Now if we multiply those slopes
together, we get three-quarters times negative four over three, which is negative
twelve over twelve, which is negative one. So it looks like those two sl-
lines would be perpendicular.
Next, which of the following lines
are parallel to each other? And then weβve got five
equations. A) π¦ equals eight π₯ minus
five. B) Two π¦ is equal to eight π₯ plus
three. C) Eight π₯ minus π¦ plus two
equals zero. D) A half π¦ minus four π₯ equals
twelve. And E) π¦ equals minus an eighth π₯
plus seven.
Well with A and E, theyβre already
in the π¦ equals ππ₯ plus π format, so itβs easy enough to read off what the slope
is. But for B, C, and D, weβre gonna
have to need to do a bit of rearranging to get them in the right format to be able
to read off their slopes. For equation B, Iβm gonna have to
divide both sides of that equation by two. So the left-hand side, half of two
π¦ is just π¦. And then dividing each term on the
right-hand side by two, half of eight π₯ is four π₯ and half of three is three over
two. For equation C, I can just add π¦
to both sides which will eliminate it from the left-hand side, and give me just π¦
on the right-hand side. So that gives me eight π₯ plus two
equals π¦. Now Iβm just gonna write this the
other way around π¦ equals eight π₯ plus two, because thatβs the format that weβre
familiar with.
Now D needs a little bit more
work. Iβve got the π¦ and the π₯ term on
the left-hand side and then just the number on the right. So first of all, Iβm gonna add four
π₯ to both sides, which gives me a half π¦ is equal to, well twelve plus four π₯ or
four π₯ plus twelve. And then doubling each side of that
equation to leave me with just π¦. Two lots of half π¦ are π¦, two
lots of four π₯ are eight π₯, and two lots of twelve are twenty-four. Now weβve got our equations in the
right format. Itβs a pretty straightforward
matter of finding the slopes, so we can see which lines are parallel to each
other. In A, the slope is eight. In B, the slope is four. In C, the slope is eight. In D, the slope is also eight. And in E, the slope is negative an
eighth. So A, C, and D, all have a slope of
eight. So the answer is A, C, and D are
parallel.
Now weβve got to write the equation
of a line which is parallel to π¦ equals eight π₯ minus four.
So itβs got to be parallel, which
means itβs got to have the same slope of eight. So itβs gotta start off with π¦
equals eight π₯. And then we can add anything we
like cause wherever that line cuts through the π¦-axis, it doesnβt matter. Itβs gonna be parallel to this
line. The only thing that you shouldnβt
use is eight π₯ minus four. Donβt make it exactly the same line
because most people would say that theyβre not parallel because itβs the same
line. So you can write anything you like
here eight π₯ plus a thousand. Here we go, thatβs a line which is
parallel to π¦ equals eight π₯ minus four.
Next, weβve got to write the
equation of a line which is perpendicular to π¦ equals three π₯ minus two.
Well, the slope is three, so the
slope of that perpendicular line must be the negative reciprocal, so thatβs minus
one over three. So our equation is gonna start off
π¦ equals minus one over three π₯. And we can add anything we
like. We could just leave it as minus one
over three π₯, or we could add any number we like to make a perpendicular line.
Now, weβve got to write an equation
for the straight line that is parallel to π¦ equals a half π₯ plus five and passes
through the point six, 10. Well, we know that the slope of the
line π¦ equals a half π₯ plus five has a slope of a half. So if our line is gonna be parallel
to that, it must also have a slope of a half. But of course, that line could cut
the π¦-axis anywhere, so we could move that line up or down. What weβre told is that it must
pass through the point six, 10. And that means that when π₯ is six,
π¦ needs to be 10.
Now if we use the general form of
our equation π¦ equals ππ₯ plus π, we know the slope π is equal to a half. Now, weβve got to find out the
value of π. But we know a particular coordinate
pair that sits on the line, when π₯ equals six, π¦ equals 10. So replacing π₯ and π¦ with six and
10, weβve got 10 is equal to a half times six plus π. So 10 is equal to three plus
π. And then subtracting three from
both sides, gives me seven is equal to π. Now, we know the value of π; we
can finish off our equation. π¦ is equal to a half π₯ plus
seven.
Lastly, write an equation for the
straight line that is perpendicular to π¦ equals three-quarters π₯ minus four and
passes through the point four, 11.
So weβre trying to find a gradient,
or a slope, which is perpendicular to three-quarters. So the slope of our perpendicular
line is gonna be negative four over three, the negative reciprocal, remember. Three-quarters times negative four
over three is equal to negative one. So that means itβs a perpendicular
line. So weβve got the slope of the line
and we also know itβs gonna go through the point four, 11. So when π₯ is equal to four, then
π¦ is equal to 11. So using the format of the equation
π¦ equals ππ₯ plus π, we know that π is minus four over three. And we know that when π₯ is four,
then π¦ is 11. So we can use all that information
to work out the value of π.
So 11 is equal to negative
four-thirds times four plus π. Well, four is the same as four over
one. So that becomes a fraction
calculation, minus four-thirds times four over one. So 11 is equal to negative
sixteen-thirds plus π. So if I add sixteen-thirds to both
sides, Iβve got 11 plus sixteen-thirds is equal to π. And that is equal to 49 over
three. So I can put that back into our
original equation and get my answer, π¦ equals negative four-thirds π₯ plus 49 over
three.