Question Video: Graphing Parallel Lines | Nagwa Question Video: Graphing Parallel Lines | Nagwa

# Question Video: Graphing Parallel Lines Mathematics • Second Year of Secondary School

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If the two straight lines 2๐ฅ + 5๐ฆ = 10 and ๐๐ฅ + ๐ฆ = 5 are parallel, which of the following graphs represents the second line?

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### Video Transcript

If the two straight lines two ๐ฅ plus five ๐ฆ equals 10 and ๐๐ฅ plus ๐ฆ equals five are parallel, which of the following graphs represents the second line?

And then weโre given five graphs to choose from. In order to help us solve this problem, letโs begin by recalling what we know about parallel lines. If these two straight lines are parallel, they will have the same slope or the same gradient. And so in order to work out which of the graphs represents the second line, weโre going to begin by finding the slope of the first line.

Now, at the moment, itโs not in a particularly useful form. So weโre going to rearrange it into slopeโintercept form, that is, of the form ๐ฆ equals ๐๐ฅ plus ๐. When itโs in this form, the value of ๐, which is the coefficient of ๐ฅ, represents the slope of the line, whereas the value of ๐ tells us the location of the ๐ฆ-intercept.

So, for the equation two ๐ฅ plus five ๐ฆ equals 10, we need to make ๐ฆ the subject. Weโll begin by subtracting two ๐ฅ from both sides. That gives us five ๐ฆ equals 10 minus two ๐ฅ. Then, we divide through by five. And of course we can individually divide each term on the right-hand side by five. 10 divided by five is two, and negative two ๐ฅ divided by five is negative two ๐ฅ over five, or negative two-fifths ๐ฅ. And so the slope of our first line is negative two-fifths. But of course since theyโre parallel, we know that the slope of our second line is also negative two-fifths.

Now, if the slope is negative, if itโs less than zero, this means that, from left to right, the line slopes downwards. Thatโs really useful because it means we can instantly disregard options (C) and (E). So, what next?

Well, letโs take the value of the slope and use it to find the value of ๐ in the equation of our second line. This will allow us to find the value of the ๐ฆ-intercept. And actually, we have three different graphs with three different ๐ฆ-intercepts. So we have the equation ๐๐ฅ plus ๐ฆ equals five. Letโs subtract ๐๐ฅ from both sides just to get it into slopeโintercept form, giving us ๐ฆ equals five minus ๐๐ฅ.

Now, the coefficient of ๐ฆ is one, so we donโt need to go any further. Since we said the slope of this second line is also negative two-fifths, the coefficient of ๐ฅ must be negative two-fifths. So negative ๐ must be equal to negative two-fifths, which means ๐ is equal to two-fifths.

Now, in fact, we see that when we put this value of ๐ into the expression ๐ฆ equals five minus ๐๐ฅ, it wasnโt actually necessary to do so to be able to work out the value of the ๐ฆ-intercept. The value of the ๐ฆ-intercept is the constant when the expression is of the form ๐ฆ equals ๐๐ฅ plus ๐. So the ๐ฆ-intercept must be equal to five. The only graph that satisfies this criteria is graph (D).

But of course itโs not enough just to find the gradient and the ๐ฆ-intercept. We do in fact need to check that this line has a gradient of negative two-fifths or, alternatively, that a second point on the line satisfies the equation ๐ฆ equals five minus two ๐ฅ.

Letโs use the second method just to verify our answer. This passes through the point with coordinates five, three. Knowing that the equation is ๐ฆ equals five minus two-fifths ๐ฅ, weโll substitute ๐ฅ equals five in. And we hope that weโll get ๐ฆ equals three back out. That gives us ๐ฆ equals five minus two-fifths times five. We can then cancel by dividing through by five. So ๐ฆ is five minus two, which is three. This means the ordered pair satisfies the equation ๐ฆ equals five minus two-fifths ๐ฅ. And this verifies to us that graph (D) is in fact the graph that we were after.

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