### Video Transcript

Find the Cartesian form of the
equation of the straight line passing through the origin and the intersection point
of the two straight lines πΏ one π equals one, one, negative two plus π‘ times one,
four, three and πΏ two π₯ equals three π¦ minus five over negative four equals π§
minus three over negative one.

Alright, so hereβs the idea. We have these two straight lines πΏ
one and πΏ two. And we know that thereβs a point
where they intersect. What we want to do is solve for the
equation of the line that passes through this intersection point as well as the
origin. Our first step in doing this will
be to figure out where πΏ one and πΏ two meet. The equation for πΏ two is given in
whatβs called symmetric form. Interpreting this form, we can say
that while π₯ is always equal to positive three, π¦ minus five over negative four
and π§ minus three over negative one are both equal to a scale factor we can call π‘
two.

This means we can write the
equation of πΏ two in whatβs called parametric form. Written this way, π₯ is equal to
three as always, while π¦, we can say, is equal to negative four times π‘ two plus
five. This is so because π¦ minus five
over negative four is equal to π‘ two. In the same way, because π§ minus
three over negative one equals π‘ two, we can say that π§ equals negative π‘ two
plus three. Now that we have our line πΏ two
written in parametric form, weβre going to do the same thing with our line πΏ
one. This line, we see, is given to us
in whatβs called vector form. If we write this equation in
parametric form, we have that π₯ is equal to one times π‘ or π‘ plus one. Then the π¦-component is equal to
four times π‘ plus one and the π§-component, three times π‘ minus two.

Now hereβs why weβve written our
two lines equations in parametric form. At the point of intersection that
we want to solve for between πΏ one and πΏ two, we can say that the π₯-values of
line two and line one are equal and the same thing for the π¦- and the
π§-values. And note that this is only at the
single point of intersection of these lines. At the single point of intersection
then, we can say that these three equations hold true. And what weβll do now is use these
equations to solve for the parameters π‘ and π‘ two.

Since there are two unknowns we
want to solve for, weβll pick two equations to use to find them out. Letβs choose the equations for the
π₯- and π¦-values of our point. Our π₯-equation, that three is
equal to π‘ plus one, implies that π‘ equals three minus one or two. If we then substitute this value
for π‘ into π‘ in our π¦-equation, we find that negative four times π‘ two plus five
equals four times two, thatβs our value for π‘, plus one.

Rearranging this equation a bit,
this means that negative four π‘ two is equal to positive four. And then if we divide both sides of
this equation by negative four, we find that π‘ two equals negative one. Now that we know the values of π‘
and π‘ two, we can substitute them into these three equations everywhere they
appear, and the values weβll solve for will be the π₯-, π¦-, and π§-coordinates of
our point of intersection of line one and line two. Doing this in our first equation
confirms that π₯ equals three where these lines meet.

In our second equation, it tells us
that the π¦-value at this point is positive nine. And from our third equation, we
find that π§ equals four. The point of intersection of our
lines, then, is three, nine, four. And now that we know this, letβs
recall a bit about the Cartesian form of the equation of a straight line. A line like this can be written in
general as π₯ minus π₯ one over π being equal to π¦ minus π¦ one over π being
equal to π§ minus π§ one over π. Here, π₯ one, π¦ one, and π§ one
are coordinates of a point that this line passes through. And π, π, and π are the
components of a vector parallel to the line.

So when it comes to the equation of
the straight line we want to solve for, if we know a point that line passes through
as well as a vector parallel to it, we can write that lineβs equation in Cartesian
form. We do already know a point that our
line passes through. Actually, we know two, the origin
and the point we solve for, three, nine, four. We do not yet know the components
of a vector parallel to our line. But note that we could solve for
such a vector by connecting up our two known points along the line. If we say that this vector has
components π, π, and π, then those values are given by the differences between
the various coordinates of our two points. We find this vector simply has
components three, nine, four.

We now have everything we need to
write out the Cartesian form of the equation of this straight line. If we choose to let the origin
correspond to the point π₯ one, π¦ one, π§ one that our line passes through, then we
can say that the equation of our line is π₯ divided by three equals π¦ divided by
nine equals π§ divided by four. Note that our final answer would
look differently if we had picked another point that lies along this line. Either way, though, the equation
would still be describing the same line in space. The Cartesian form weβll report as
our final answer then is π₯ over three equals π¦ over nine equals π§ over four.