Video: AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 1 β€’ Question 27

AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 1 β€’ Question 27

05:50

Video Transcript

The points 𝐴, 𝐡, and 𝐢 are points on the circumference of a circle. And 𝐡𝐢 is the diameter of the circle. The coordinates of 𝐴 are three, two and the coordinates of 𝐡 are zero, zero. Work out the equation of the line 𝐴𝐢.

We’ll begin just by adding the coordinates of points 𝐴 and 𝐡 onto our diagram. So we have three, two for 𝐴 and zero, zero for 𝐡. We’re asked to work out the equation of the line 𝐴𝐢.

So we can recall that the equation of any straight line can be written in the form 𝑦 equals π‘šπ‘₯ plus 𝑐, where π‘š represents the gradient of the line and 𝑐 represents the 𝑦-intercept. Notice that this is a lowercase 𝑐 representing the 𝑦-intercept. It’s not the same and shouldn’t be confused with the point 𝐢 in this diagram.

So to find the equation of the line 𝐴𝐢, we need to work out the values of π‘š and 𝑐. The gradient of the line joining two points can be found by dividing their change in 𝑦 by their change in π‘₯.

If we think of the points as having coordinates π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two, then this can be formalized as 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. Now, this looks complicated. But in fact, all it’s saying is we need to find the difference between the 𝑦-values and then divide it by the difference in the π‘₯-values.

It doesn’t matter which way around we subtract the 𝑦-coordinates and which way around we subtract the π‘₯-coordinates. We’ll get the same result in either case, as long as we’re consistent with what we do for π‘₯ and what we do for 𝑦.

Now, we don’t know the coordinates of the point 𝐢, which means we can’t work out the gradient of the line 𝐴𝐢. But we do know the coordinates of 𝐴 and 𝐡. So we could work out the gradient of this line. We’ll see how that helps us in a moment.

If we subtract the coordinates of 𝐡 from those of 𝐴 then, then we have two minus zero for the change in 𝑦 and three minus zero for the change in π‘₯. Two minus zero is two and three minus zero is three. So the gradient of the line 𝐴𝐡 is two-thirds.

Now, how is this useful to us? Well, the key point is that 𝐡𝐢 is the diameter of the circle, which means that it divides circle into two semicircles. One of our key circle theorems is that the angle in a semicircle or the angle inscribed in a semicircle is a right angle. It’s equal to 90 degrees.

So this means that angle 𝐢𝐴𝐡 is 90 degrees. This also tells us that the line 𝐴𝐡 is perpendicular to the line 𝐴𝐢. The reason this is useful is because we also know that the gradients of perpendicular lines multiply to negative one.

If we think of the two gradients as π‘š one and π‘š two, then we can write this as π‘š one π‘š two. That means π‘š one multiplied by π‘š two equals negative one. But if we divide both sides of this equation by one of the gradients, then we also have that π‘š one is equal to negative one over π‘š two.

This means that if we know the gradient of one of the lines, we can work out the gradient of the other line by dividing negative one by the known gradient. This is also called finding a negative reciprocal.

So to find the gradient of 𝐴𝐢, we divide negative one by two-thirds. To divide by a fraction, we flip or invert that fraction and we multiply. So this is equivalent to negative one multiplied by three over two, which gives negative three over two.

Now, in fact, there is a little shortcut that we could have taken here. If we want to find the negative reciprocal of a fraction, we can actually just flip or invert that fraction and change its sign. So the negative reciprocal of π‘Ž over 𝑏 is negative 𝑏 over π‘Ž.

Notice that this is the case here. We flip the fraction two-thirds to form three over two and change the value from positive to negative. So substituting this value of negative three over two as the gradient π‘š in the equation of our line gives 𝑦 equals negative three over two π‘₯ plus 𝑐.

And we just need to determine the value of 𝑐. To do so, we need to substitute the coordinates of any point that lies on the line into the equation and then solve the resulting equation for 𝑐.

The only point that lies on the line whose equation we know is the point 𝐴, which has coordinates three, two. So we substitute two for 𝑦 and three for π‘₯. And it gives two equals negative three over two multiplied by three plus 𝑐.

To solve this equation, we first remember that three can be thought of as the fraction of three over one. So we’re multiplying negative three over two by three over one. And remembering that we multiply the numerators together and multiply the denominators together, this simplifies to negative nine over two.

We can also write the integer two as the fraction four over two because four divided by two is two. So now, we have a common denominator for the numbers on either side of this equation.

To solve for 𝑐, we need to add nine over two to each side of the equation, giving four over two plus nine over two equals 𝑐. As these fractions have a common denominator, we can just add the numerators. And we see that 𝑐 is equal to 13 over two.

Finally, we just need to substitute this value of 𝑐 into the equation of our line. And we have that the equation of the line 𝐴𝐢 is 𝑦 equals negative three over two π‘₯ plus 13 over two.

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