Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 2 β€’ Question 23

A line 𝐿₁ is parallel to the line 𝐿₂ and passes through the point (7, βˆ’2). The equation of 𝐿₂ is 2𝑦 + 4π‘₯ = 5. Find the equation of the line 𝐿₁.

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Video Transcript

A line 𝐿 one is parallel to the line 𝐿 two and passes through the point seven, negative two. The equation of 𝐿 two is two 𝑦 plus four π‘₯ equals five. Find the equation of the line 𝐿 one.

Let’s recall the general equation for a straight line. It’s 𝑦 equals π‘šπ‘₯ plus 𝑐, where π‘š is the gradient of the line β€” that tells us how steep it is β€” and 𝑐 is the value of the 𝑦-intercept β€” that tells us where the line crosses the 𝑦-axis. To find the equation of 𝐿 one then, we’re going to need to find its gradient and its 𝑦-intercept. We’re told that lines 𝐿 one and 𝐿 two are parallel. For two lines to be parallel, that means they must have the same gradient, the same steepness. So if we can find the gradient of the line 𝐿 two, then we’ll also have the gradient of the line 𝐿 one. We can’t however easily find the gradient of the line 𝐿 two yet, since it’s not in the same form as our general equation of a straight line.

A common mistake is to think that the gradient is always the coefficient of π‘₯, the number of π‘₯s there are. In this case, that’s four. But we can’t do that until our equation is in the form 𝑦 equals π‘šπ‘₯ plus 𝑐. Let’s begin then by rearranging our equation to make 𝑦 the subject. The opposite of adding four π‘₯ is to subtract four π‘₯, and we’re going to need to take them away from both sides of the equation. That tells us that two 𝑦 is equal to negative four π‘₯ plus five. We’re then going to divide both sides of this equation by two. Two 𝑦 divided by two is 𝑦. Negative four π‘₯ divided by two is negative two π‘₯. And we aren’t particularly worried about the value of the 𝑦-intercept of this line, however five divided by two is 2.5.

We can now see that the coefficient of π‘₯, the number of π‘₯s we have here, is negative two. And therefore, the gradient of both our lines is negative two. Let’s substitute this into the general equation of a straight line. If we do that, we get 𝑦 is equal to negative two π‘₯ plus 𝑐. Notice how we haven’t yet used the information in the question, that the line passes through the point seven, negative two. Remember these coordinates represent an π‘₯ value and a 𝑦 value. On this line, when π‘₯ is seven, 𝑦 is equal to negative two. So we can substitute these values into our equation. 𝑦 is negative two, and negative two π‘₯ is negative two multiplied by seven plus 𝑐. Negative two multiplied by seven is negative 14.

We’re nearly there. We need to solve this equation for 𝑐. And we’re going to do that by adding 14 to both sides. If we start at negative two on the number line and add 14, that tells us to go up the number line, and we end up at 12. So 𝑐 is equal to 12. We can now take this value of 𝑐 and put it back into the equation that we started with. And we get 𝑦 is equal to negative two π‘₯ plus 12. Now we are finished at this point, but we could check our work by going back and substituting the values of π‘₯ is seven and 𝑦 is negative two into this equation. Negative two is equal to negative two multiplied by seven plus 12. That tells us that negative two is equal to negative two, which we know to be true, and it’s certain that we’ve done the very last bit at least completely correct. 𝑦 is equal to negative two π‘₯ plus 12.

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