Video: Pack 1 β€’ Paper 2 β€’ Question 19

Pack 1 β€’ Paper 2 β€’ Question 19

03:39

Video Transcript

A quadrilateral has vertices at the point 𝐴, 𝐡, 𝐢, and 𝐷, which have coordinates negative two, two; two, six; three, three; and one, one, respectively. 𝐸 is the midpoint of 𝐴𝐡. And 𝐹 is the midpoint of 𝐢𝐷. Prove that lines 𝐴𝐡 and 𝐸𝐹 are perpendicular. You must explain every step of your working out.

Two lines are perpendicular if their gradients, given here as π‘š one and π‘š two, have a product of negative one. In this case then, we need to work out the gradients of the lines 𝐴𝐡 and 𝐸𝐹. Remember, the formula for gradient is change in 𝑦 divided by change in π‘₯, which is sometimes written as 𝑦 two minus 𝑦 one all over π‘₯ two minus π‘₯ one.

To find the gradient of the line 𝐴𝐡, we therefore need to substitute the coordinates of 𝐴 and 𝐡 into this formula. Change in 𝑦 is six minus two. And change in π‘₯ is two minus negative two.

Remember, it’s absolutely fine to perform this calculation the other way round as long as you’re consistent. If you do two minus six for change in 𝑦, you then have to do negative two minus two for change in π‘₯. Six minus two is four. And two minus negative two, which becomes two plus two, is also four. Four divided by four is one. The gradient of the line passing between 𝐴 and 𝐡 is, therefore, one.

Now before we can work out the gradient of the line 𝐸𝐹, we first need to work out the coordinates of both 𝐸 and 𝐹. 𝐸 is the midpoint of 𝐴𝐡. And 𝐹 is the midpoint of 𝐢𝐷. To find the midpoints, we add together the π‘₯-coordinates and divide by two and we add together the 𝑦-coordinates and divide by two.

Remember, 𝐸 is the midpoint of 𝐴 and 𝐡. So its π‘₯-coordinate is negative two plus two all over two. And its 𝑦-coordinate is two plus six, again all over two. Negative two plus two is zero. So dividing it by two, we still get zero. And two plus six divided by two is eight divided by two, which is four.

We can find the coordinates for 𝐹 by repeating this process for points 𝐢 and 𝐷. In this case, three plus one over two is the π‘₯-coordinate and three plus one over two is also the 𝑦-coordinate. Three plus one is four, which divided by two is two. So the coordinate for 𝐹 is two, two.

Once we have these coordinates, we can then find the gradient of the line passing through 𝐸𝐹 using the formula from before. Change in 𝑦 is two minus four. And change in π‘₯ is two minus zero. Two minus four is negative two. And negative two divided by two is negative one. We found the gradient of the line 𝐸𝐹 to be negative one.

Now remember, we said that two lines are perpendicular if the product, which means multiply, of their gradients is negative one. The gradient of 𝐴𝐡 is one and the gradient of 𝐸𝐹 is negative one. One multiplied by negative one is negative one. Don’t forget to include a conclusion. 𝐴𝐡 and 𝐸𝐹 are perpendicular.

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