Chapter 4 · Geometry and Equations · Concept 1
4-4 Points Equidistant from Two Points
Two towns and one road. Where along the road is the point that is exactly as far from one town as from the other? Every question in this lesson is that question, dressed differently.
Guiding question
How do you find a point when all you know is what it is equally far from?
Learning outcome. Find the position of a point equidistant from two points.
Name the point with ONE letter
You are always told which line the point sits on. Spend that fact immediately, by writing the point with only one unknown in it:
on the x-axis
(s, 0)
y is 0 all along it
on the y-axis
(0, t)
x is 0 all along it
on y = ax + b
(s, as + b)
the y follows from the x
Why one letter and not two.
“Equally far from A and B” gives you exactly one equation. With two unknowns you would be stuck. The line you were given is the second piece of information, and writing (s, 2s − 1) is how you spend it before you start.
And squaring both sides is safe here. Both distances are positive, so PA = PB and PA² = PB² say exactly the same thing — which means the roots can come off straight away and never come back.
All the points equally far from A and B lie on one straight line. Your answer is just where that line crosses the one you were given — which is why there is always exactly one.
the point on the x-axis that is the same distance from A(-2, 1) and B(3, 4)
2. Find Q on y = 2x − 1 that is the same distance from A(1, 4) and B(3, 1).
- 1
Name Q with one letter, using the line you were given.
Q(s, 2s − 1)Every point on that line has its y equal to 2x − 1, so calling the x s fixes the y as well.
Work with a partner. After you have an answer, both of you work out the two distances separately. They must come out the same — and if they do not, you know at once rather than in the exam.
Two of these have fraction answers.
5/6 and 17/2 are perfectly good coordinates — the crossing point simply does not land on a grid line. A fraction here is not a sign that something has gone wrong.
A(2, 1), B(4, 5), C(3, y) and D(x, 0), with CA = CB and DA = DB. Find the distance CD.
Using two letters for the point.
Write (s, 0) or (s, 2s − 1), not (s, t). The line you were given is a fact — spend it before you start, or you will have one equation and two unknowns.
Stopping at s.
The question asks for a POINT. Once you have s = 13/8, put it back into the line to get the y as well.
Expecting two answers.
The s² terms always cancel, so the equation is linear and there is exactly one point. If you end up with a quadratic, something has been expanded wrongly.
Losing a sign in the y-part.
For A(4, −5) and a point (s, 3s + 1), the difference is (3s + 1) + 5 = 3s + 6. Subtracting a negative adds.
Distrusting a fraction.
5/6 and 17/2 are correct answers. The crossing point does not have to land on a grid line.
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