### Video Transcript

If the distances of the point 𝑃
with coordinates 𝑥, 𝑦 from 𝐴 with coordinates five, one and 𝐵 with coordinates
negative one, five are equal, prove that three 𝑥 equals two 𝑦.

We can draw a diagram to help
us. Where is the point 𝐴? Well, it has coordinates five comma
one. So it’s here. How about 𝐵? Well, it has coordinates negative
one, five. And so it’s here. The point 𝑃 is an arbitrary point
in the plane with coordinates 𝑥 and 𝑦. All we know is that the distances
of 𝑃 from 𝐴 and 𝐵 are both the same. So let’s mark in our point
somewhere. Let’s put it here, for
example. Let’s mark the coordinates in. Now, we want these distances to be
the same. This diagram is just to help us
understand what we have to prove. It’s not needed as part of the
proof itself. But, of course, it couldn’t
hurt.

Okay, so what are these distances
that we want to be equal? The distance of 𝑃 from 𝐴 is
𝐴𝑃. And the distance of 𝑃 from 𝐵 is
𝐵𝑃. To find these distances, we use the
distance formula. The distance formula tells us that
the distance between two points with coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two
is the square root of 𝑥 two minus 𝑥 one squared plus 𝑦 two minus 𝑦 one
squared.

To find the distance between the
point with coordinates five, one and the point with coordinates 𝑥, 𝑦, we
substitute five, one for 𝑥 one, 𝑦 one and 𝑥, 𝑦 for 𝑥 two, 𝑦 two in our
formula. We get the square root of the
difference of the 𝑥-coordinates squared plus the difference of the 𝑦-coordinates
squared. It’s exactly the same for 𝐵𝑃. We have the difference of the
𝑥-coordinates squared. And, of course, 𝑥 minus negative
one is just 𝑥 plus one, plus the difference of the 𝑦-coordinates squared.

Okay, we found the distances of 𝑃
from 𝐴 and 𝐵 using the distance formula. Now, we can set these two distances
equal to find a relation between the coordinates 𝑥 and 𝑦 of 𝑃. Before we do this, let me just
remark that had we forgotten the distance formula, we could find the distances of 𝑃
from 𝐴 and 𝐵 from first principles. To find the distance of 𝑃 from 𝐴,
that’s the length of the line segment 𝐴𝑃, we can consider this right-angled
triangle which makes 𝐴𝑃 the hypotenuse. Looking at the 𝑥 coordinates of 𝐴
and 𝑃, we can see that the base has a length of five minus 𝑥. And the height is the difference in
the 𝑦-coordinates, 𝑦 minus one.

Applying Pythagoras’s theorem, we
get the length of the hypotenuse 𝐴𝑃, as before. The only possible difference is
that we might have five minus 𝑥 squared instead of 𝑥 minus five squared. But these are equal. Okay, so however we got the
distances 𝐴𝑃 and 𝐵𝑃, we’re told that they’re equal. So we should set them equal to each
other. 𝐴𝑃 equals 𝐵𝑃. We substitute the expressions we
found for these distances.

We can square both sides and then
expand the brackets. For example, 𝑥 minus five squared
is 𝑥 squared minus two times five 𝑥 plus five squared. This is using the fact that 𝑥
minus 𝑎 squared is 𝑥 squared minus two 𝑎𝑥 plus 𝑎 squared. We can simplify this and move on to
the next term. We expand 𝑦 minus one squared to
𝑦 squared minus two 𝑦 plus one. On the right-hand side, we expand
𝑥 plus one squared to 𝑥 squared plus two 𝑥 plus one and 𝑦 minus five squared to
𝑦 squared minus 10𝑦 plus 25.

Having done this, we see that there
is some cancellation. An 𝑥 squared on the left-hand side
cancels with an 𝑥 squared on the right. And similarly, the 𝑦 squareds on
the left- and right-hand sides also cancel. Combining the constant terms 25 and
one on the left-hand side, we get negative 10𝑥 minus two 𝑦 plus 26 on the
left-hand side. And combining the constant terms on
the right-hand side, we get two 𝑥 minus 10𝑦 plus 26. The constant terms cancel. And we’re left with only 𝑥- and
𝑦-terms. Adding 10𝑥 to both sides, we’re
left with just negative two 𝑦 on the left-hand side. And the 10𝑥 combines with the two
𝑥 on the right-hand side to give 12𝑥 minus the 10𝑦.

We can also add 10𝑦 to both sides
to get 10 minus two equals eight 𝑦 on the left-hand side and just 12𝑥 on the
right-hand side. We can divide both sides by four,
getting two 𝑦 on the left-hand side and three 𝑥 on the right-hand side. And finally, swapping both sides we
get that three 𝑥 is equal to two 𝑦, which is what we were required to prove.

You might like to think about what
this means graphically. The equation three 𝑥 equals two 𝑦
defines a line on the coordinate plane. We have proved that if a point 𝑃
is equidistant from points 𝐴 and 𝐵, then it lies on this line. You might recognize this line,
whose equation we found, as the perpendicular bisector of the line segment 𝐴𝐵. This question used the distance
formula which is really worth remembering. But you can, of course, rederive it
if you’ve forgotten it using Pythagoras’s theorem. Once we applied the distance
formula, the rest was just algebra.