### Video Transcript

The graph shows a line connecting
π to a point π. The gradient of the line is
one-third. The length of the line is the
square root of 1210. Find the coordinates of the point
π.

As weβre asked to find the
coordinates of this point π, letβs give them some letters to work with. We can refer to the coordinates of
π as π, π. If we sketch in a right-angled
triangle below this line, then we know that if the coordinates of π are π, π,
then the length of the base of this triangle, the horizontal side, is π units. Thatβs the π₯-coordinate of π. And the length of the vertical side
of the triangle, the height, is π units. Thatβs the π¦-coordinate of π. We can then form some equations
using the two pieces of information weβve been given about this line.

Firstly, weβre told that the
gradient of this line is one-third. Now the definition of the gradient
of a line is the change in π¦ divided by the change in π₯. You might also see this written as
rise over run, which just means the vertical change over the horizontal change. From our graph, the rise is π and
the run is π. So we can form an equation. π over π is equal to one-third,
as that is what we were told the gradient to the line is equal to.

We can simplify this equation by
cross-multiplying, that is, multiplying both sides by both denominators. When we do this, the π in the
denominator on the left cancels out. And now we have three π. And on the right, the three in the
denominator cancels out. And weβre left with π. So our equation simplifies to three
π is equal to π. We can label this as equation
one.

The other piece of information
weβve been given is that the length of the line is the square root of 1210. As this triangle is a right-angled
triangle, we can apply Pythagorasβs theorem to find another relationship between π
and π.

Pythagorasβs theorem tells us that,
in a right-angled triangle, the sum of the squares of the two shorter sides is equal
to the square of the longest side, the hypotenuse. And if we label the two shorter
sides as π and π, as they are here, and the longest side as π, we have π squared
plus π squared equals π squared.

So substituting the square root of
1210 for π, we have that π squared plus π squared equals the square root of 1210
squared. Now if we square-root 1210 and then
square it, these two operations cancel each other out. So weβre just left with 1210. The equation becomes π squared
plus π squared equals 1210. And this we can label as equation
two.

Weβre now going to solve equations
one and two simultaneously to find the values of π and π. And as equation one tells us that
π is equal to three π, weβre going to substitute this expression for π into
equation two. Doing so gives three π all squared
plus π squared equals 1210.

So now weβve eliminated π and we
have an equation in terms of π only. Weβll expand the brackets around
three π all squared. And we need to be careful. Weβre squaring the three as well as
the π. So three π all squared is equal to
nine π squared, not three π squared. Watch out for this common
mistake.

Next, we can collect the like terms
on the left-hand side of the equation. Nine π squared plus π squared
gives 10π squared. And then we can divide both sides
of the equation by 10 to give that π squared is equal to 121. Next, we take the square root of
each side of the equation, giving that π is equal to plus or minus the square root
of 121, which is plus or minus 11.

But notice from our diagram that π
is in the first quadrant, which means that both its π₯- and π¦-coordinates are
positive. Usually, we would need to write
plus or minus the square root when solving an equation by square rooting. But in this case, we just want the
positive answer. So we have that π is equal to
11.

Now that we found the value of π,
we can substitute this back into equation one in order to find the value of π. π, remember, is equal to three
π. So thatβs three multiplied by 11,
which is equal to 33. The coordinates of point π then,
which are π, π, are 33, 11.