Video: Determining the Shortest Distance between a Line and a Point

Find the shortest distance between the point (βˆ’6, 10) and the line which passes through the points (1, 9) and (4, 6).

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Video Transcript

Find the shortest distance between the point negative six, 10 and the line which passes through the points one, nine and four, six.

Well, in order to find the shortest distance between a point and a line, then this is gonna be the perpendicular distance. So therefore, what we want to use is use the formula for this. But to enable us to use the formula, what we first need to do is have the equation of the straight line in the form π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 equals zero. And also, we need to have the coordinates of the point that we’re trying to find the distance from that point to the straight line. And they’re in the form π‘₯ sub one, 𝑦 sub one. And once we have this, then we have the formula 𝐿, which is our perpendicular or shortest distance between a point and a line, is equal to the modulus or absolute value of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐 all over the square root of π‘Ž squared plus 𝑏 squared.

So therefore, in our context, the first thing we need to do is find out the equation of the straight line and then put it into the form π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 equals zero. But what we want to do is use the general form for the equation of a straight line which is 𝑦 equals π‘šπ‘₯ plus 𝑐, where π‘š is the slope and 𝑐 is the 𝑦-intercept. Well, we can do that, but the first thing we want to do is work out the slope of our line. Well, to find the slope of our line, then what this is is the change in 𝑦 divided by the change in π‘₯. So therefore, we could say that π‘š, our slope, is equal to 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one.

Well, in the straight line we’re looking at, we know that it passes through two points: one, nine and four, six. So therefore, we’ve got our π‘₯ sub one, 𝑦 sub one; π‘₯ sub two, 𝑦 sub two. So, what we’ve got is π‘š is equal to six minus nine over four minus one. So therefore, π‘š, our slope, is gonna be equal to negative three over three, which is gonna give us a slope of negative one. Okay, great! So we can plug this into our general form for the equation of a straight line. So when we do that, what we got is 𝑦 equals negative π‘₯ plus 𝑐.

Well, now, to find out what 𝑐 is, we can substitute either of our pairs of coordinates into the equation 𝑦 equals negative π‘₯ plus 𝑐. So I’ve chosen the one which is the point one, nine. So therefore, if we use this, we get nine is equal to negative one plus 𝑐. So then, if we add one to each side of the equation, we’re gonna get 𝑐 is equal to 10. So what we could do now is substitute this back in, and we got the equation of our straight line. So when we do that, we get 𝑦 equals negative π‘₯ plus 10. So then, we want to rearrange this into the form π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 equals zero. We could just subtract 𝑦 from each side of the equation. Or, just to keep it neat and keep the first term positive, what I’m gonna do is add π‘₯ and subtract 10 from each side of the equation. So when we do that, we get π‘₯ plus 𝑦 minus 10 equals zero.

So when we do this, we’ve got our π‘Ž-, 𝑏-, and 𝑐-values. So they’re gonna be π‘Ž is one, 𝑏 is one, and 𝑐 is negative 10. And then for this part of the question, our π‘₯ sub one is equal to negative six and our 𝑦 sub one is equal to 10 because what these are are the coordinates of the point that we’re looking to find the distance from that point to the straight line. So when we substitute in our values, we get 𝐿 is equal to the modulus or absolute value of one multiplied by negative six plus one multiplied by 10 minus 10 all over the square root of one squared plus one squared. So what we’re gonna get is 𝐿 is equal to the modulus or absolute value of negative six over root two.

So therefore, we’re gonna get 𝐿 is equal to six over root two because if you got the modulus or absolute value of negative six, well we just want the magnitude. So we just want the positive value, so we get six over root two. Well, now, because we’ve got a surd on the denominator, what we want to do is rationalize the denominator, so we remove a surd from the denominator. And to do that, we multiply by root two over root two. And we do that because root π‘Ž multiplied by root π‘Ž is equal to π‘Ž, so therefore root two multiplied by root two is just gonna be equal to two. So this is gonna give us 𝐿 is equal to six root two over two. So then, what we can do is divide the numerator and the denominator by two. When we do that, we get 𝐿 is equal to three root two.

So we can say that the shortest distance between the point negative six, 10 and the line which passes through the points one, nine and four, six is three root two.

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