### Video Transcript

What is the distance between the point negative nine, negative 10 and the line of slope one through three, negative seven?

So in this question, what weβre trying to find is the shortest distance or the distance between the point and our line, and what this can also be known as is the perpendicular distance. And for this we have a formula. However, for our formula to work, what we need is the equation of our line to be in the form ππ₯ plus ππ¦ plus π equals zero. And also we have to have our point, which is π₯ sub one, π¦ sub one. And if we have this, then the equation weβre gonna use is πΏ, our perpendicular distance, is equal to the modulus or absolute value of ππ₯ sub one plus ππ¦ sub one plus π all divided by the square root of π squared plus π squared.

Well, first of all, what we need to do, as we said, is get the equation of our straight line, and we want it in the form ππ₯ plus ππ¦ plus π equals zero. Well, to find the equation of our straight line, what we have is general form of the equation of a straight line, which is π¦ equals ππ₯ plus π, where π is the slope and π is the π¦-intercept. So therefore, we can say that the equation of our line is gonna be π¦ equals π₯ plus π, and thatβs because our slope or π is equal to one. So weβve got π¦ equals one π₯ plus π, but we donβt write the one. So now what we need to do is find out what π is, so we need to find out what our π¦-intercept is. But how are we going to do that?

Well, to find our π, what weβre gonna do is substitute values of π₯ and π¦ that we have from the point that we know is on our line, and that is three and negative seven. So we know that π₯ is gonna be equal to three and π¦ is equal to negative seven. So when we do that, we get negative seven is equal to three plus π. So then, if we subtract three from each side of the equation, weβre gonna get negative 10 equals π. So then, if we substitute this back into the equation of our straight line, weβre gonna get π¦ is equal to π₯ minus 10. But this still isnβt quite in the form we wanted because it isnβt in the form ππ₯ plus ππ¦ plus π equals zero. So what we need to do now is rearrange so that it is.

So therefore, if we subtract π¦ from each side of the equation, what weβre gonna get is zero equals π₯ minus π¦ minus 10. So now we have our π, π, and π cause π is gonna be equal to one, π is equal to negative one, and π is equal to negative 10. So now that we have our π, π, and π, we can also see that weβve got our π₯ sub one and π¦ sub one because what these are are the coordinates of the point that weβre looking for, the distance from this point to the line. So theyβre gonna be π₯ sub one is equal to negative nine and π¦ sub one is equal to negative 10. So weβve got all the values we need, so we can substitute these into our formula. So then weβre gonna have πΏ is equal to the modulus or absolute value of one multiplied by negative nine plus negative one multiplied by negative 10 minus 10 and then all divided by one squared plus negative one squared.

So therefore, weβre gonna have πΏ is equal to the absolute value or modulus of negative nine over root two, which is just gonna give nine over root two because if we got the absolute value or modulus, that means we just want the magnitude or positive value. Now, what we do here because weβve got a surd as the denominator is we rationalize the denominator. And to do that, what we do is multiply by root two over root two. But why does this work? Why do we do this? Well, we do this because root π multiplied by root π is just π. So therefore, root two multiplied by root two is gonna be two, so it means weβre not gonna have a surd as our denominator. So therefore, what weβre gonna get is that πΏ is equal to nine root two over two.

So we can say that the distance between the point negative nine, negative 10 and the line of slope one through three, negative seven is, as we said, nine root two over two.