Orienting Yourself: The Use of Coordinates · Lesson 4 of 5
Distance Between Two Points in the 2-D Plane
“Because even points on a plane occasionally need to know how far apart they are.”
• Find distances between points on horizontal or vertical lines. • Understand horizontal and vertical coordinate differences as perpendicular displacements. • Derive the general distance formula using the Baudhāyana–Pythagoras theorem. • Use the distance formula when coordinates are positive or negative. • Understand why reflections preserve distances.
Coordinates tell us where points are. The next question is: how far apart are two points? If the points lie on a horizontal or vertical line, the answer comes directly from a coordinate difference. If they lie diagonally, we can build a right triangle and use the Baudhāyana–Pythagoras theorem.
Horizontal Distance
Suppose P(x₁, y) and Q(x₂, y) have the same y-coordinate. They lie on a horizontal line. Their distance is the magnitude of the horizontal change.
Problem
Find the distance between P(−3, 4) and Q(5, 4).
- 1.The y-coordinates are equal, so the segment is horizontal.
- 2.Distance = |5 − (−3)|.
- 3.Distance = |8| = 8 units.
Vertical Distance
Suppose P(x, y₁) and Q(x, y₂) have the same x-coordinate. They lie on a vertical line. Their distance is the magnitude of the vertical change.
Problem
Find the distance between A(2, −5) and B(2, 3).
- 1.The x-coordinates are equal, so the segment is vertical.
- 2.Distance = |3 − (−5)|.
- 3.Distance = 8 units.
When the Segment Is Diagonal
For a diagonal segment, subtracting only the x-coordinates or only the y-coordinates does not give the true straight-line distance. Those two differences give the horizontal and vertical shifts. Since these shifts are perpendicular, they can be treated as the legs of a right triangle.
A Concrete Example Before the Formula
Problem
Find AD.
- 1.Horizontal change = 7 − 3 = 4 units.
- 2.Vertical change = 4 − 1 = 3 units in magnitude.
- 3.These two changes form perpendicular sides of a right triangle.
- 4.By the Baudhāyana–Pythagoras theorem, AD² = 4² + 3².
- 5.AD² = 16 + 9 = 25.
- 6.Therefore AD = 5 units.
This example reveals the general pattern. The horizontal leg comes from the difference of x-coordinates, the vertical leg comes from the difference of y-coordinates, and the required distance is the hypotenuse.
Deriving the General Distance Formula
Let A(x₁, y₁) and B(x₂, y₂) be any two points. Construct F(x₁, y₂). Then AF is vertical and FB is horizontal, so triangle AFB is right-angled at F.
Applying the Baudhāyana–Pythagoras theorem gives the square of the hypotenuse as the sum of the squares of the two perpendicular legs.
Taking the positive square root gives the distance. We take the positive root because distance is never negative.
Why Negative Differences Do Not Cause Trouble
The coordinate differences may be positive or negative depending on which point is subtracted first. But the differences are squared in the formula. Squaring removes the sign, so the final distance is unchanged.
Problem
Find the distance between A(−3, 4) and D(−7, 1).
- 1.x₂ − x₁ = −7 − (−3) = −4.
- 2.y₂ − y₁ = 1 − 4 = −3.
- 3.Distance = √[(-4)² + (-3)²].
- 4.Distance = √(16 + 9) = 5 units.
- 5.The negative coordinate differences do not create a negative distance because they are squared.
Finding All Sides of a Triangle
Problem
For A(3,4), D(7,1) and M(9,6), find AD, DM and MA.
- 1.AD = √[(7−3)² + (1−4)²] = √25 = 5.
- 2.DM = √[(9−7)² + (6−1)²] = √29.
- 3.MA = √[(3−9)² + (4−6)²] = √40.
- 4.So the three side lengths are 5, √29 and √40 units.
Reflection and Distance
Reflecting a point in the y-axis changes the sign of its x-coordinate but keeps its y-coordinate unchanged. For example, (3,4) becomes (−3,4). The reflected figure is not stretched or compressed, so corresponding lengths remain the same.
| Original point | Reflection in y-axis |
|---|---|
| A(3,4) | A′(−3,4) |
| D(7,1) | D′(−7,1) |
| M(9,6) | M′(−9,6) |
Problem
Compare AD for A(3,4), D(7,1) with A′D′ for A′(−3,4), D′(−7,1).
- 1.AD = √[(7−3)² + (1−4)²] = √25 = 5.
- 2.A′D′ = √[(−7−(−3))² + (1−4)²].
- 3.A′D′ = √[(-4)² + (-3)²] = √25 = 5.
- 4.The side length is preserved by the reflection.
Using Distance to Test Geometric Claims
The distance formula can answer geometric questions without depending only on a drawing. We can test whether several points are equally far from the origin, compare the side lengths of a shape, or investigate whether three points lie on one straight line by comparing the three pairwise distances.
Problem
Check whether M(−3,−4), A(0,0) and G(6,8) lie on one straight line using distances.
- 1.MA = √[(0+3)² + (0+4)²] = 5.
- 2.AG = √[(6−0)² + (8−0)²] = 10.
- 3.MG = √[(6+3)² + (8+4)²] = 15.
- 4.MA + AG = 5 + 10 = 15 = MG.
- 5.Therefore A lies between M and G, so the three points are collinear.
Practice Problems
- Find the distance between (−4,3) and (5,3).
- Find the distance between (2,−6) and (2,7).
- Find the distance between A(2,1) and B(−1,2).
- For A(2,1), B(−1,2), C(−2,−1) and D(1,−2), calculate all side lengths and both diagonals. What type of quadrilateral do the distances suggest?
- Check using distances whether R(−5,−1), B(−2,−5) and C(4,−13) are collinear.
- A circle is centred at O(0,0) with radius √65. Decide whether P(1,−8), Q(−4,7), D(−5,6) and E(0,9) lie inside, on or outside the circle.
Key Takeaways
• Horizontal distance is |x₂−x₁| and vertical distance is |y₂−y₁|. • A diagonal segment can be studied through a right triangle. • The distance formula follows directly from the Baudhāyana–Pythagoras theorem. • Negative differences do not affect the result because they are squared. • Distances can be used to test geometric properties without relying only on a drawing.
Next, we revise the complete coordinate system and practise location, quadrants, distances and coordinate-based reasoning together.