Finding Distance Between Points - Sixth Grade Mathematics

Calculating the distance between points that share a common coordinate line develops spatial reasoning and absolute value applications on the coordinate plane.

Horizontal and Vertical Alignment

In sixth grade mathematics, we focus on finding the exact distance between two points that lie on the same horizontal line (sharing identical y-coordinates) or on the same vertical line (sharing identical x-coordinates).

Horizontal Line Segment:     y-coordinates are IDENTICAL.
                             Example: A(-3, 4) and B(5, 4)  --> Same y = 4
                             Distance = | x1 - x2 |

Vertical Line Segment:       x-coordinates are IDENTICAL.
                             Example: C(2, -5) and D(2, 3)  --> Same x = 2
                             Distance = | y1 - y2 |

The Distance Calculation Protocols

There are two primary spatial cases when calculating distance: points lying in the same quadrant, and points lying in different quadrants (crossing an axis).

+-------------------------------------------------------------------------+
|                  DISTANCE BETWEEN ALIGNED POINTS PROTOCOL               |
+-------------------+-----------------------+-----------------------------+
| Spatial Position  | Computational Method  | Concrete Example            |
+-------------------+-----------------------+-----------------------------+
| Same Quadrant     | SUBTRACT the smaller  | Points: (2, 8) and (2, 3)   |
| (Same Sign)       | absolute value from   | Distance = |8| - |3|        |
|                   | larger absolute value | = 8 - 3 = 5 units           |
+-------------------+-----------------------+-----------------------------+
| Different Quads   | ADD the absolute      | Points: (-4, 2) and (5, 2)  |
| (Different Signs /| values of the         | Distance = |-4| + |5|       |
| Crossing an Axis) | differing coordinates | = 4 + 5 = 9 units           |
+-------------------+-----------------------+-----------------------------+

Case 1: Same Sign (Same Quadrant)

When two points share an axis alignment and their differing coordinates have the same sign (both positive or both negative), they lie on the same side of the axis. The distance is found by subtracting their absolute values: Points (3, 7) and (3, 2): Distance = 7 - 2 = 5 units. Points (-2, -8) and (-2, -3): Distance = |-8| - |-3| = 8 - 3 = 5 units.

Case 2: Different Signs (Crossing an Axis)

When two points have differing coordinates with opposite signs (one positive and one negative), the line segment crosses an axis. To find the total distance, add their individual distances from the axis: Points A(-4, 3) and B(6, 3): Point A is 4 units to the left of the y-axis (|-4| = 4), and Point B is 6 units to the right of the y-axis (|6| = 6). The total distance is 4 + 6 = 10 units.

Visualizing Crossing the y-axis between (-4, 3) and (6, 3):
<---|---|---|---|---|---|---|---|---|---|---|--->
   -4  -3  -2  -1   0   1   2   3   4   5   6
    |<- 4 units ->| |<------- 6 units ------>|
    Total Distance = 4 + 6 = 10 units!

The Universal Absolute Value Difference Formula

Both cases can be expressed using a single universal formula: For horizontal segments: Distance = | x1 - x2 |. For vertical segments: Distance = | y1 - y2 |. Because distance is an absolute difference, the order in which you subtract does not matter: |-4 - 6| = |-10| = 10, and |6 - (-4)| = |6 + 4| = |10| = 10.

Exercises

Exercise 1. Find the distance between each pair of points: (a) (4, 9) and (4, 2); (b) (-3, -7) and (-3, -1); (c) (-5, 6) and (8, 6); (d) (2, -8) and (2, 7).

Exercise 2. A town map is plotted on a coordinate grid where each unit represents 1 mile. The town hall is at (-4, 3) and the library is at (6, 3). How far is the library from the town hall?

Exercise 3. Point P is located at (5, -3). Point Q is located on the same vertical line, 9 units away from Point P. What are the two possible coordinates for Point Q?

Solutions

Solution to Exercise 1. For (a) (4, 9) and (4, 2): x-coordinates are identical (4); vertical segment: distance = |9 - 2| = 7 units. For (b) (-3, -7) and (-3, -1): x-coordinates are identical (-3); vertical segment in same quadrant: distance = |-7 - (-1)| = |-7 + 1| = |-6| = 6 units. For (c) (-5, 6) and (8, 6): y-coordinates are identical (6); horizontal segment crossing y-axis: distance = |-5| + |8| = 5 + 8 = 13 units. For (d) (2, -8) and (2, 7): x-coordinates are identical (2); vertical segment crossing x-axis: distance = |-8| + |7| = 8 + 7 = 15 units.

Solution to Exercise 2. The town hall (-4, 3) and library (6, 3) share the same y-coordinate (3). The distance is horizontal across the y-axis: Distance = |-4| + |6| = 4 + 6 = 10 units. Since each unit represents 1 mile, the library is exactly 10 miles from the town hall.

Solution to Exercise 3. Point P is at (5, -3). Because Point Q is on the same vertical line, its x-coordinate must also be 5. Being 9 units away means Point Q could be 9 units above P, or 9 units below P. Moving 9 units upward: -3 + 9 = +6, giving Q(5, 6). Moving 9 units downward: -3 - 9 = -12, giving Q(5, -12). The two possible coordinates for Point Q are (5, 6) and (5, -12).