Question Video: Finding the Equation of a Straight Line | Nagwa Question Video: Finding the Equation of a Straight Line | Nagwa

Question Video: Finding the Equation of a Straight Line Mathematics • Third Year of Preparatory School

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Write, in the form 𝑦 = π‘šπ‘₯ + 𝑐, the equation of the line through 𝐴(5, βˆ’8) that is perpendicular to the line 𝐴𝐡, where 𝐡(βˆ’8, βˆ’3).

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Video Transcript

Write in the form 𝑦 equals π‘šπ‘₯ plus 𝑐, the equation of the line through point 𝐴 which is five, negative eight that is perpendicular to 𝐴𝐡, where point 𝐡 is negative eight, negative three.

Well, the first thing we have got is this word here perpendicular cause what we want to do is find the equation of the line through point 𝐴 that is perpendicular to 𝐴𝐡. Well, there is a relationship with perpendicular lines that’s gonna help us to solve this problem. And this relationship involves their slopes. Because what we can say is if we have two lines, one which is perpendicular to the other β€” so we’ll call them π‘š one and π‘š two. This is the slopes, so slope one, slope two. If you multiply the slopes together, you get negative one.

So therefore, what we can say is if we divided by one of our slopes, then we could say that β€” for instance, here we’ve got π‘š one is equal to negative one over π‘š two. And what this means is it’s actually the negative reciprocal. So, we can say that if lines are perpendicular, then their slopes are the negative reciprocal of one another.

And to give you an example of how this’d work, if we had one slope that was equal to a third, then the slope of the other line, which I’ve said is perpendicular, would be equal to negative three over one, which would just be negative three. Cause with the reciprocal, if we have a fraction like a third, then we swap the numerator and the denominator. And as we’ve said before, it’s got to be negative. Okay, so, we know the relationship between a line and the line that’s perpendicular to it.

And as we’ve already discussed, we’re gonna be working with slopes. And to help us find the slope, what we have is a formula. And that formula is that π‘š, our slope, is equal to 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. So, what this means is the change in 𝑦 divided by the change in π‘₯.

Well, if we take a look at our points A and 𝐡, so we could now work out the slope of 𝐴𝐡, the line 𝐴𝐡. So, I’ve labelled our different coordinates π‘₯ one, 𝑦 one, π‘₯ two, 𝑦 two. So therefore, if we substitute in our values for π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two, we can have the slope of 𝐴𝐡 β€” So, I’ll put this as π‘š 𝐴𝐡 β€” is equal to negative three minus negative eight over negative eight minus five. Remembering that if we subtract a negative, it’s an add. Be careful of this because negative numbers can often cause mistakes.

So, now, we got negative three add eight over negative eight minus five. Which is gonna give us five over negative 13. So, we can say that the slope of 𝐴𝐡 is negative five over 13. So therefore, we can say the slope of the perpendicular β€” so, that’s the line that we’re looking to find the equation of β€” is going to be the negative reciprocal of negative five over 13. So therefore, it’s going to be 13 over five.

And if we think, well, would this give us our negative one if we multiplied them together? Well, if we multiply negative five over 13 and 13 over 5, we get five multiplied by 13 as the numerator, so that goes 65. And then, five multiplied by 13 on the denominator, which will be 65. So, that would be negative 65 over 65, which would be our negative one.

So, now, we have the slope. What we want to do is work out the equation of the line through 𝐴 that is perpendicular to the line 𝐴𝐡. So, to do that, we’re gonna use 𝑦 equals π‘šπ‘₯ plus 𝑐. So, 𝑦 equals π‘šπ‘₯ plus 𝑐 is the general form for an equation of a straight line where π‘š, as we’ve already mentioned, is the slope and 𝑐 is the 𝑦-intercept.

So, as you can see, we’ve got 𝑦 equals 13 over five π‘₯ plus 𝑐 because we know our π‘š. So, this is okay. But we don’t know what 𝑐 is; we don’t know what our 𝑦-intercept is. So, how are we going to find that? To help us find 𝑐, we’ve got a point that we know that the line goes through cause we know that the line goes through point 𝐴, which is five, negative eight.

So therefore, if we substitute in our π‘₯- and 𝑦-values, we’re gonna get negative eight plus our 𝑦 is equal to 13 over five multiplied by five then plus 𝑐. So, this is gonna give us negative eight is equal to 13 plus 𝑐. And that’s because if you have 13 over five multiplied by five, it’s just 13 because the fives would cancel.

So, then, if we subtract 13 from each side of the equation, we’re gonna get negative 21 is equal to 𝑐. So therefore, if we substitute this value in to the equation we had in 𝑦 equals π‘šπ‘₯ plus 𝑐. We can say that the equation of the line through the point 𝐴, which is five, negative eight, that is perpendicular to the line A𝐡 where 𝐡 is equal to negative eight, negative three. Then, 𝑦 is gonna be equal to 13 over five π‘₯ minus 21.

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