# Video: AQA GCSE Mathematics Higher Tier Pack 3 β’ Paper 2 β’ Question 26

The graph shows a line connecting π to a point π. The gradient of the line is 1/3. The length of the line is β1210. Find the coordinates of the point π.

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### 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 weβre asked to find the coordinates of this point π, letβs 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. Thatβs the π₯-coordinate of π. And the length of the vertical side of the triangle, the height, is π units. Thatβs the π¦-coordinate of π. We can then form some equations using the two pieces of information weβve been given about this line.

Firstly, weβre 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 weβre 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 weβve 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 Pythagorasβs theorem to find another relationship between π and π.

Pythagorasβs 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 weβre just left with 1210. The equation becomes π squared plus π squared equals 1210. And this we can label as equation two.

Weβre 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, weβre going to substitute this expression for π into equation two. Doing so gives three π all squared plus π squared equals 1210.

So now weβve eliminated π and we have an equation in terms of π only. Weβll expand the brackets around three π all squared. And we need to be careful. Weβre 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 thatβs three multiplied by 11, which is equal to 33. The coordinates of point π then, which are π, π, are 33, 11.