We're asked to determine x,
which is the x-coordinate of the second ordered pair
so that the distance between the two points is five units.
The distance formula for two points on the coordinate plane
is shown below, where the distance is equal
to the square root of the square of x sub two
minus x of one plus the square of y sub two minus y sub one,
where the first ordered pair is x sub one comma y sub one,
and the second ordered pair is x sub two comma y sub two.
So, in our case, let's go and label our ordered pairs.
We have x sub one comma y sub one,
x two comma two y sub two.
Notice how we're looking for the x-coordinate
of the second ordered pair
so that the distance between the two points is five units.
Let's substitute all the known values
into the distance formula.
We know the distance D is five,
giving us five equals the square root.
The radicand is the square of x sub two minus x sub one,
plus the square of y sub two minus y sub one,
which gives us the square of x minus two,
plus the square of two minus negative one.
And now, we need to simplify, and then solve for x.
Well, two minus negative one
simplifies to two plus one, or three,
giving us five equals the square root
of the square of x minus two, plus the square of three.
And now, let's undo the square root on the right
by squaring both sides of the equation.
Remember squaring will undo the square root.
So, on the left, five squared is equal to 25.
On the right, again, squaring undoes a square root,
leaving us with the square of x minus two.
Plus the square of three, the square of three is nine.
The next step is a square x minus two.
To do this, we have two factors of x minus two.
When multiplying two binomials, we have four products.
We distribute x, and then we distribute negative two.
We have 25 equals x times x is x squared,
and we have x times negative two plus negative two times x.
That's negative two x plus two x, or negative four x,
giving us minus four x.
And then, negative two times negative two is positive four,
giving us plus four.
We still have plus nine.
Four plus nine is 13, giving us 25 equals x squared
minus four x plus 13.
Because we have a quadratic equation,
we will set the equation equal to zero,
and hopefully, solve by factoring.
Set the left side equal to zero,
we subtract 25 on both sides.
Simplifying, we have zero equals x squared minus four x.
13 minus 25 is negative 12.
If the right side does factor,
it will factor into two binomial factors.
The factors of x squared are x and x.
Because the coefficient of x squared is equal to one,
the constants of the binomial factors
will be the factors of negative 12
that add to negative four.
Well, negative six times positive two is negative 12,
and negative six plus two is negative four,
and therefore, we have a factor of x minus six
and a factor of x plus two.
Notice here we have two solutions.
Notice that the product on the right is equal to zero
when x minus six is equal to zero,
or when x plus two is equal to zero.
So, we will have two solutions for x.
We have x equals six or x equals negative two.
So, going back up to the problem,
we now know the distance between two comma negative one
and six comma two is equal to five units, but also,
or the distance between two comma negative one
and negative two comma two is also five units.
Again, we found the x-coordinate of the second ordered pair
could be positive six or negative two.
So, we have two possible solutions.
Before we go, though, let's take a look at this
on the coordinate plane.
The given point was two common negative one.
This point here, and then we found x equals six
or x equals negative two.
When x equals six, six comma two
is this point here on the right.
The length of the blue segment is five units.
We also found x equals negative two,
which means a second point
could also be negative two comma two,
which is graphed here on the right.
The length of the green segment is also five units.
I hope you found this helpful.