Question Video: Determining the 𝑦-Coordinates for Two Points Lying on Two Parallel Lines | Nagwa Question Video: Determining the 𝑦-Coordinates for Two Points Lying on Two Parallel Lines | Nagwa

# Question Video: Determining the π¦-Coordinates for Two Points Lying on Two Parallel Lines Mathematics • Third Year of Preparatory School

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If the line passing through points π΄ (β13, 8) and π΅ (20, π¦) is parallel to the line passing through points πΆ (β2, 0) and π· (7, π¦), what is the value of π¦?

03:26

### Video Transcript

If the line passing through points π΄: negative thirteen, eight and π΅: twenty, π¦ is parallel to the line passing through points πΆ: negative two, zero and π·: seven, π¦, what is the value of π¦?

For two lines to be parallel, they must have the exact same slope. They must be equal. So in order for the slopes to be the same, we would need to set the slope formulas equal to each other for each line.

So to find the slope of a line, itβs the change in π¦ divided by the change in π₯, so π¦ two minus π¦ one over π₯ two minus π₯ one. Itβs also commonly written as π¦ one minus π¦ two over π₯ one minus π₯ two.

As long as you follow that same pattern, either formula would be fine. Here we have our points. π΄ and π΅ create a line as well as πΆ and π·; they create a line. Now weβll label our points. π΄ the negative thirteen is π₯ one and eight is the π¦ one and π΅ thatβs our second point thatβs listed, so twenty is π₯ two and π¦ is our π¦ two, and same thing with πΆ and π·.

Letβs first begin by finding the slope of π΄π΅, so π¦ two minus π¦ one, so π¦ minus eight, over π₯ two minus π₯ one, so twenty minus negative thirteen. So really thatβs adding.

So on the top, we have π¦ minus eight, which doesnβt simplify. And on the bottom, twenty plus thirteen is equal to thirty three. Now letβs do the same thing for line πΆπ·.

So for slope of πΆπ·, we would take π¦ two minus π¦ one, so π¦ minus zero, over π₯ two minus π₯ one, so seven minus negative two. So the slope of πΆπ· will be π¦ minus zero all over nine.

Now these lines are parallel, which means they should have the exact same slope. So we can take both of these fractions and set them equal to each other, and we can take these again because these lines are supposed to be parallel. So the slope should be equal.

So if we set them equal to each other, we can find the cross product and solve for π¦. So here weβre setting our slopes equal to each other and now we will cross-multiply.

So we have nine times π¦ minus eight equal to thirty three times π¦ minus zero. Now we distribute. So we will take nine times π¦ and then nine times negative eight and thirty three times π¦ and thirty three times zero.

So we have nine π¦ minus seventy two equals thirty three π¦. The minus zero doesnβt really do anything. So now we need to isolate the π¦. So letβs go ahead and add nine π¦ to both sides of the equation.

Doing so, the nine π¦s cancel. Weβve negative seventy two left on the left-hand side of the equation. And on the right-hand side, thirty three π¦ plus nine π¦ is twenty four π¦.

Now our last step would be to divide both sides by twenty four, which means π¦ is equal to negative three. So the value of π¦ knowing that these lines are parallel, we take the slopes, we set them equal to each other, and then we solved for π¦. And we got that π¦ is equal to negative three.

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