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

π΄π΅πΆπΈ is a trapezium. π· is a point on the line πΆπΈ. The ratio of the line πΈπ· : π·πΆ = 1 : 7. Given that π is perpendicular to the vector π, show that π΄πΈ = π΅πΆ.

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### Video Transcript

π΄π΅πΆπΈ is a trapezium. π· is a point on the line πΆπΈ. The ratio of the line πΈπ· to π·πΆ is equal to one to seven. Given that π is perpendicular to the vector π, show that π΄πΈ is equal to π΅πΆ.

To show that π΄πΈ is equal to π΅πΆ, 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 weβll look at how to find the length of the line in a moment.

We will however begin by finding the vectors that describe π΄ to πΈ and π΅ to πΆ. There is no vector to describe the journey that takes us from π΄ to πΈ. Instead, weβll need to find an alternative labelled route.

We can begin by travelling from π΄ to π΅. Then weβll travel from π΅ to π· and finally from π· to πΈ. In vector terms, we can say that the vector π΄πΈ is equal to the vector π΄π΅ plus the vector π΅π· plus the vector π·πΈ. Notice though how the vector π΅π· is travelling against the direction labelled. To change the direction of a vector, we change its sign. Thatβs a bit like multiplying by negative one. Similarly, the vector π· to πΈ is travelling against the direction marked for πΈπ·. So we change its sign.

Letβs add the given vectors into this formula. The vector π΄π΅ is four π. We then subtract the vector π·π΅, which is five π plus four π, and we subtract the vector πΈπ·, which is π. Four π minus five π minus π is negative two π. And we end up with negative two π minus four π.

Letβs now find the vector that describes the journey from π΅ to πΆ. Once again, thereβs no labelled journey. So instead, we travel from π΅ to π· and then from π· to πΆ. We can see from this ratio that the length π·πΆ is seven times the length of πΈπ·. Itβs travelling in the same direction as πΈπ·, so we can say that π·πΆ is seven π.

Once again, we remember that the vector π΅π· is the same as negative π·π΅. And our expression becomes negative five π plus four π plus seven π. Negative five π plus seven π is two π. And we end up with two π minus four π for the vector π΅πΆ.

We now need to find the lengths of the lines π΄πΈ and π΅πΆ. We know that π΄ is perpendicular to π΅. Letβs imagine that π΄ is travelling in the horizontal direction and π΅ is travelling in the vertical direction. If this was the case, we could say that π΄πΈ is moving two units left and four units down, whereas π΅πΆ 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 Pythagorasβs theorem to prove this. We add a third line showing the actual journey taken. This is the hypotenuse. Itβs the line thatβs opposite the right angle in each of these triangles.

Pythagorasβs theorem says that the sum of the squares of the two smaller sides is equal to the square of the longer side. Itβs often written as π squared plus π squared equals π squared, where π is the hypotenuse.

Substituting π and π for two and four in this first triangle, we get two squared plus four squared equals π squared. Two squared is four, and four squared is 16. So we get π squared is equal to four plus 16, which is 20. We solve this equation by finding the square root of both sides. So π 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 π is equal to root 20. And weβve shown that the length of π΄πΈ and the length of π΅πΆ is equal. In fact, we didnβt need to draw these triangles. We could have used a formula.

The length of π΄πΈ is equal to the square root of the sum of the squares of the coefficients of π and π. Thatβ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 π΄πΈ.

For π΅πΆ, this does look slightly different, since weβre not actually using triangles. Weβre using the coefficients of π and π. 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 π΅πΆ. And we have shown that the lengths π΄πΈ and π΅πΆ are root 20, and π΄πΈ is equal to π΅πΆ.