### Video Transcript

If the point 𝐴 at zero, two is
equidistant from the points 𝐵 at three, 𝑝 and 𝐶 at 𝑝, five, find 𝑝, and then
find the length of 𝐴𝐵.

We are told that the point 𝐴 is
equidistant from points 𝐵 and 𝐶. We can therefore use the distance
formula to create an equation in terms of 𝑝. The distance formula says that for
two points 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, the distance between them is found by
the square root of the square of the differences between the 𝑥-coordinates plus the
square of the differences between the 𝑦-coordinates.

Let’s begin by forming an
expression for the length of the line segment 𝐴𝐵. Now it doesn’t really matter what
we choose to be 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two as long as we’re consistent. In this case, choosing 𝐴 to be 𝑥
one, 𝑦 one and 𝐵 to be 𝑥 two, 𝑦 two gives us that the length or the distance
between them is the square root of three minus zero squared plus 𝑝 minus two
squared. Three minus zero squared is
nine. So we get that the length 𝐴𝐵 is
the square root of nine plus 𝑝 minus two squared.

We’ll repeat this process for the
length of the line segment 𝐴𝐶. That gives us the square root of 𝑝
minus zero squared plus five minus two squared. 𝑝 minus zero squared is simply 𝑝
squared. And five minus two squared is three
squared which is nine. So the length of the line segment
𝐴𝐶 is the square root of 𝑝 squared plus nine.

Since 𝐴𝐵 and 𝐴𝐶 are
equidistant, this means that 𝐴𝐵 and 𝐴𝐶 are of equal length. We can therefore equate these two
expressions to get an equation in terms of 𝑝. We can first find the square of
both sides to get nine plus 𝑝 minus two all squared equals 𝑝 squared plus
nine. We can then subtract nine from both
sides. And we get 𝑝 minus two all squared
equals 𝑝 squared. We’ll need to expand these brackets
before we go any further.

Be careful, a common mistake is to
think that we just square the 𝑝 and the two and we’re done. In fact, we’ll need to consider
this as a product of two brackets and expand from there. That is to say we’ll need to write
it as 𝑝 minus two multiplied by 𝑝 minus two. Multiplying the first term in the
first bracket by the first term in the second bracket gives us 𝑝 squared. 𝑝 multiplied by negative two is
negative two 𝑝. And negative two multiplied by 𝑝
is once again negative two 𝑝. In fact, we can write negative two
𝑝 minus two 𝑝 as negative four 𝑝. Finally, we multiply negative two
by negative two to give us four. And we now have 𝑝 squared minus
four 𝑝 plus four is equal to 𝑝 squared.

Notice that we can subtract 𝑝
squared from both sides. And that gives us negative four 𝑝
plus four is equal to zero. To solve this, we’ll first add four
𝑝 to both sides to give us four equals four 𝑝. And then we’ll divide through by
four. That gives us that 𝑝 is equal to
one. Remember, 𝑝 is a coordinate. So we don’t need to include any
units. We do, however, still need to find
the length of 𝐴𝐵.

Remember, we created an expression
for the length of 𝐴𝐵 in terms of 𝑝. So we can now substitute one into
this expression. Doing so, and we get that the
length of 𝐴𝐵 is the square root of nine plus one minus two all squared. One minus two is negative one
squared. And negative one squared is
one. So the length becomes the square
root of nine plus one. The length of 𝐴𝐵 is root 10
units.