Video: AQA GCSE Mathematics Foundation Tier Pack 1 โ€ข Paper 1 โ€ข Question 16

Consider the line with equation 2๐‘ฅ + 2๐‘ฆ = 4. The points ๐พ and ๐ฟ lie on the line. (a) Complete the coordinates of ๐พ and ๐ฟ. ๐พ(0, ๏ผฟ) ๐ฟ(๏ผฟ, 0)(b) Hence or otherwise, draw the line 2๐‘ฅ + 2๐‘ฆ = 4 for the values of ๐‘ฅ from โˆ’3 to 3.

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Video Transcript

Consider the line with equation two ๐‘ฅ plus two ๐‘ฆ equals four. The points ๐พ and ๐ฟ lie on the line. Part a) Complete the coordinates of ๐พ and ๐ฟ. Part b) Hence or otherwise, draw the line two ๐‘ฅ plus two ๐‘ฆ equals four for the values of ๐‘ฅ from negative three to three.

So weโ€™ve been given the equation of this line. Two ๐‘ฅ plus two ๐‘ฆ equals four. And weโ€™re told that the points ๐พ and ๐ฟ lie on the line. Now if a point lies on a line, then it means that the coordinates of that point satisfy the equation of that line. So for any point on this line, it is the case that two multiplied by its ๐‘ฅ-coordinate plus two multiplied by its ๐‘ฆ-coordinate is always equal to four.

Letโ€™s consider the point ๐พ first of all. We know that its ๐‘ฅ-coordinate is zero. But we donโ€™t know its ๐‘ฆ-coordinate. So we can just call it ๐‘ฆ. As ๐พ lies on the line, we can substitute its coordinates into the equation of the line. Two ๐‘ฅ becomes two multiplied by zero. Two ๐‘ฆ is just still two ๐‘ฆs. We donโ€™t know the value of ๐‘ฆ. And this is equal to four.

We now have an equation that we can solve in order to find the value of ๐‘ฆ. Two multiplied by zero is just zero. So our equation becomes two ๐‘ฆ is equal to four. To solve for ๐‘ฆ, we need to divide both sides of this equation by two. Two ๐‘ฆ divided by two is ๐‘ฆ, and four divided by two is two. So we have that ๐‘ฆ is equal to two. And therefore, the ๐‘ฆ-coordinate of the point ๐พ is two.

For the point ๐ฟ, we know that the ๐‘ฆ-coordinate is zero. And itโ€™s the ๐‘ฅ-coordinate that weโ€™re looking to find this time. We can do this in exactly the same way. Two ๐‘ฅ is still two ๐‘ฅ. And then two ๐‘ฆ becomes two multiplied by zero. So we have two ๐‘ฅ plus two multiplied by zero is equal to four. Two multiplied by zero is still zero. So we have two ๐‘ฅ equals four. And to solve for ๐‘ฅ, we need to divide both sides of the equation by two, giving ๐‘ฅ equals two. The coordinates of the point ๐ฟ then are two, zero.

Now letโ€™s think about how this helps us with part b, which is asking us to draw the line two ๐‘ฅ plus two ๐‘ฆ equals four for values of ๐‘ฅ between negative three and three. We know that this is a straight line because there is a linear relationship between ๐‘ฅ and ๐‘ฆ. The only power of ๐‘ฅ and ๐‘ฆ that appears in the equation is one. We donโ€™t have any ๐‘ฅ squareds or ๐‘ฆ squared terms, for example.

We also know the coordinates of two points that lie on the line. We have ๐พ, which is the point zero, two. We can plot this. Zero on the ๐‘ฅ-axis and two on the ๐‘ฆ-axis gives us the point here. We also know that the point ๐ฟ lies on our line, which has coordinates two, zero. So we go two units to the right of the origin and no units up. And it gives us the second point, here.

As this is a straight line, we only need two points in order to be able to draw it. We can now take our ruler and connect the two points that weโ€™ve drawn and then extend this line. However, itโ€™s always good to have at least three points to make sure we havenโ€™t made any mistakes.

We can choose any ๐‘ฅ-value that we like, or indeed any ๐‘ฆ-value, to work out the corresponding ๐‘ฆ- or ๐‘ฅ-value. Letโ€™s choose ๐‘ฅ equals negative three. We substitute into the equation of the line in exactly the same way as we did in part a.

We have two multiplied by ๐‘ฅ โ€” thatโ€™s two multiplied by negative three โ€” plus two ๐‘ฆ is equal to four. Two multiplied by negative three is negative six. Remember, positive multiplied by a negative gives a negative answer. So we have negative six plus two ๐‘ฆ is equal to four.

To solve for ๐‘ฆ, our next step is to add six to each side of the equation, giving two ๐‘ฆ is equal to 10. We can then divide both sides of the equation by two to give ๐‘ฆ is equal to five. This means that the third point on our line has the coordinates negative three, five. And so we can also plot this point.

We now take our ruler and connect the three points weโ€™ve drawn with a straight line, noting that they do indeed all lie on the same line. As weโ€™ve been asked to draw the line for ๐‘ฅ-values from negative three to three, we must make sure that our line reaches at least these values. But we can extend our line beyond these points if we wish.

The word โ€œhenceโ€ in a question like this means we need to use the work weโ€™ve just done. So we use the coordinates of the points ๐พ and ๐ฟ, which weโ€™d worked out in part a, to help us draw this line.

The word โ€œotherwiseโ€ suggests that we could actually do this a different way. What we could do is rearrange the equation of our straight line into the general form ๐‘ฆ equals ๐‘š๐‘ฅ plus ๐‘. Here ๐‘š gives the gradient of the straight line and ๐‘ gives the ๐‘ฆ-intercept, the value at which the line crosses the ๐‘ฆ-axis. We could use these two key features in order to draw the line.

Our line has equation two ๐‘ฅ plus two ๐‘ฆ is equal to four. We would first subtract two ๐‘ฅ from each side to give two ๐‘ฆ equals negative two ๐‘ฅ plus four and then divide the equation by two to give ๐‘ฆ equals negative ๐‘ฅ plus two. Comparing this with the general equation of a straight line, we see that the gradient of our line is negative one, because the coefficient or value in front of the ๐‘ฅ is negative one, and the ๐‘ฆ-intercept is positive two.

The ๐‘ฆ-intercept is this point that Iโ€™ve now marked in pink on our graph, the point where the line crosses the ๐‘ฆ-axis. And we can see that it is indeed two on the line that weโ€™ve drawn. The gradient of the line tells us, for every one unit we move to the right, how many units the line moves up or down. So the gradient of negative one means that, for every one unit we move to the right, the line goes down by one unit, which we can see is indeed the case on the line weโ€™ve drawn.

Our alternative method then wouldโ€™ve been to draw in the ๐‘ฆ-intercept at two and then draw some other points by moving one unit right and one unit down each time. We could then connect these points with a straight line.

So weโ€™ve completed the question. In part a, we found that the coordinates of ๐พ and ๐ฟ was zero, two and two, zero, respectively. And then in part b, weโ€™ve discussed two possible methods for drawing the line with equation two ๐‘ฅ plus two ๐‘ฆ is equal to four.

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