Video: AQA GCSE Mathematics Higher Tier Pack 3 β€’ Paper 2 β€’ Question 27

𝐴𝐡𝐢𝐸 is a trapezium. 𝐷 is a point on the line 𝐢𝐸. The ratio of the line 𝐸𝐷 : 𝐷𝐢 = 1 : 7. Given that π‘Ž is perpendicular to the vector 𝑏, show that 𝐴𝐸 = 𝐡𝐢.

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Video Transcript

𝐴𝐡𝐢𝐸 is a trapezium. 𝐷 is a point on the line 𝐢𝐸. The ratio of the line 𝐸𝐷 to 𝐷𝐢 is equal to one to seven. Given that π‘Ž is perpendicular to the vector 𝑏, show that 𝐴𝐸 is equal to 𝐡𝐢.

To show that 𝐴𝐸 is equal to 𝐡𝐢, we do not need to show that their vectors are the same. Instead, we need to show that the magnitudes of their vectors are the same. This just means their length. And we’ll look at how to find the length of the line in a moment.

We will however begin by finding the vectors that describe 𝐴 to 𝐸 and 𝐡 to 𝐢. There is no vector to describe the journey that takes us from 𝐴 to 𝐸. Instead, we’ll need to find an alternative labelled route.

We can begin by travelling from 𝐴 to 𝐡. Then we’ll travel from 𝐡 to 𝐷 and finally from 𝐷 to 𝐸. In vector terms, we can say that the vector 𝐴𝐸 is equal to the vector 𝐴𝐡 plus the vector 𝐡𝐷 plus the vector 𝐷𝐸. Notice though how the vector 𝐡𝐷 is travelling against the direction labelled. To change the direction of a vector, we change its sign. That’s a bit like multiplying by negative one. Similarly, the vector 𝐷 to 𝐸 is travelling against the direction marked for 𝐸𝐷. So we change its sign.

Let’s add the given vectors into this formula. The vector 𝐴𝐡 is four π‘Ž. We then subtract the vector 𝐷𝐡, which is five π‘Ž plus four 𝑏, and we subtract the vector 𝐸𝐷, which is π‘Ž. Four π‘Ž minus five π‘Ž minus π‘Ž is negative two π‘Ž. And we end up with negative two π‘Ž minus four 𝑏.

Let’s now find the vector that describes the journey from 𝐡 to 𝐢. Once again, there’s no labelled journey. So instead, we travel from 𝐡 to 𝐷 and then from 𝐷 to 𝐢. We can see from this ratio that the length 𝐷𝐢 is seven times the length of 𝐸𝐷. It’s travelling in the same direction as 𝐸𝐷, so we can say that 𝐷𝐢 is seven π‘Ž.

Once again, we remember that the vector 𝐡𝐷 is the same as negative 𝐷𝐡. And our expression becomes negative five π‘Ž plus four 𝑏 plus seven π‘Ž. Negative five π‘Ž plus seven π‘Ž is two π‘Ž. And we end up with two π‘Ž minus four 𝑏 for the vector 𝐡𝐢.

We now need to find the lengths of the lines 𝐴𝐸 and 𝐡𝐢. We know that 𝐴 is perpendicular to 𝐡. Let’s imagine that 𝐴 is travelling in the horizontal direction and 𝐡 is travelling in the vertical direction. If this was the case, we could say that 𝐴𝐸 is moving two units left and four units down, whereas 𝐡𝐢 is two units right and four units down.

Instinctively, it kind of makes sense that these will form the same length lines. But we can actually use Pythagoras’s theorem to prove this. We add a third line showing the actual journey taken. This is the hypotenuse. It’s the line that’s opposite the right angle in each of these triangles.

Pythagoras’s theorem says that the sum of the squares of the two smaller sides is equal to the square of the longer side. It’s often written as π‘Ž squared plus 𝑏 squared equals 𝑐 squared, where 𝑐 is the hypotenuse.

Substituting π‘Ž and 𝑏 for two and four in this first triangle, we get two squared plus four squared equals 𝑐 squared. Two squared is four, and four squared is 16. So we get 𝑐 squared is equal to four plus 16, which is 20. We solve this equation by finding the square root of both sides. So 𝑐 in our first triangle is equal to root 20.

We can see in our second triangle that the equation is exactly the same. Once again, we see that 𝑐 is equal to root 20. And we’ve shown that the length of 𝐴𝐸 and the length of 𝐡𝐢 is equal. In fact, we didn’t need to draw these triangles. We could have used a formula.

The length of 𝐴𝐸 is equal to the square root of the sum of the squares of the coefficients of π‘Ž and 𝑏. That’s a negative two squared plus negative four squared. Negative two squared is still four, and negative four squared is still 16. So we still get root 20 for the length of 𝐴𝐸.

For 𝐡𝐢, this does look slightly different, since we’re not actually using triangles. We’re using the coefficients of π‘Ž and 𝑏. Here we get two squared plus negative four squared. Two squared is once again four though, and negative four squared is still 16. So we still get root 20 for 𝐡𝐢. And we have shown that the lengths 𝐴𝐸 and 𝐡𝐢 are root 20, and 𝐴𝐸 is equal to 𝐡𝐢.

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