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 we’re 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 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero. And also we have to have our point, which is 𝑥 sub one, 𝑦 sub one. And if we have this, then the equation we’re gonna use is 𝐿, our perpendicular distance, is equal to the modulus or absolute value of 𝑎𝑥 sub one plus 𝑏𝑦 sub one plus 𝑐 all divided by the square root of 𝑎 squared plus 𝑏 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 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 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 𝑦 equals 𝑚𝑥 plus 𝑐, where 𝑚 is the slope and 𝑐 is the 𝑦-intercept. So therefore, we can say that the equation of our line is gonna be 𝑦 equals 𝑥 plus 𝑐, and that’s because our slope or 𝑚 is equal to one. So we’ve got 𝑦 equals one 𝑥 plus 𝑐, but we don’t write the one. So now what we need to do is find out what 𝑐 is, so we need to find out what our 𝑦-intercept is. But how are we going to do that?
Well, to find our 𝑐, what we’re gonna do is substitute values of 𝑥 and 𝑦 that we have from the point that we know is on our line, and that is three and negative seven. So we know that 𝑥 is gonna be equal to three and 𝑦 is equal to negative seven. So when we do that, we get negative seven is equal to three plus 𝑐. So then, if we subtract three from each side of the equation, we’re gonna get negative 10 equals 𝑐. So then, if we substitute this back into the equation of our straight line, we’re gonna get 𝑦 is equal to 𝑥 minus 10. But this still isn’t quite in the form we wanted because it isn’t in the form 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero. So what we need to do now is rearrange so that it is.
So therefore, if we subtract 𝑦 from each side of the equation, what we’re gonna get is zero equals 𝑥 minus 𝑦 minus 10. So now we have our 𝑎, 𝑏, and 𝑐 cause 𝑎 is gonna be equal to one, 𝑏 is equal to negative one, and 𝑐 is equal to negative 10. So now that we have our 𝑎, 𝑏, and 𝑐, we can also see that we’ve got our 𝑥 sub one and 𝑦 sub one because what these are are the coordinates of the point that we’re looking for, the distance from this point to the line. So they’re gonna be 𝑥 sub one is equal to negative nine and 𝑦 sub one is equal to negative 10. So we’ve got all the values we need, so we can substitute these into our formula. So then we’re gonna have 𝐿 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, we’re gonna have 𝐿 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 we’ve 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 𝑎 multiplied by root 𝑎 is just 𝑎. So therefore, root two multiplied by root two is gonna be two, so it means we’re not gonna have a surd as our denominator. So therefore, what we’re gonna get is that 𝐿 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.