Tenth Grade Mathematics: The Distance Formula and Geometric Verifications

Measuring the physical straight-line distance between two locations is the most elementary operation in geometry. When points are defined purely by their numerical coordinates in the Cartesian plane, the distance between them is governed directly by the Pythagorean theorem. The Distance Formula translates the horizontal and vertical coordinate separations into the exact straight-line Euclidean distance, establishing an algebraic tool to verify collinearity, classify geometric polygons, and establish properties of circles.

Derivation of the Distance Formula

Let P(x1, y1) and Q(x2, y2) be any two arbitrary points in the Cartesian plane. We construct a right-angled triangle PQR whose sides are parallel to the coordinate axes:

Geometric Construction of Right Triangle PQR:
                        y
                        |                      Q(x2, y2)
                        |                      +
                        |                     /|
                        |                    / |  Vertical distance
                        |        Distance d /  |  = |y2 - y1|
                        |                  /   |
                        |      P(x1, y1)  +----+ R(x2, y1)
                        |                 Horizontal distance
                        |                 = |x2 - x1|
                        +--------------------------------- x

Draw a horizontal line segment from P parallel to the x-axis, and a vertical line segment from Q parallel to the y-axis. Let these two perpendicular lines intersect at point R. The coordinates of R are (x2, y1). The length of the horizontal leg PR is the horizontal separation: PR = |x2 - x1|. The length of the vertical leg QR is the vertical separation: QR = |y2 - y1|. Triangle PQR is a right-angled triangle with the right angle at R.

By the Pythagorean theorem:

PQ^2 = PR^2 + QR^2
PQ^2 = (x2 - x1)^2 + (y2 - y1)^2

Taking the principal (positive) square root gives the universal Distance Formula:

Theorem (The Distance Formula):
The distance d between any two points P(x1, y1) and Q(x2, y2) is given by:
                 d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

Notice that because (x2 - x1)^2 = (x1 - x2)^2, the order in which the two points are subtracted does not affect the resulting distance.

Corollary: Distance from the Origin:
The distance of any point P(x, y) from the origin O(0, 0) is:
                    OP = sqrt( x^2 + y^2 )

Geometric Verifications Using the Distance Formula

The distance formula allows us to prove geometric classifications purely through side-length computations:

+-----------------------+-------------------------------------------------------------+
| Geometric Figure      | Condition to Prove Using Distance Formula                   |
+-----------------------+-------------------------------------------------------------+
| Collinear Points      | Sum of two smaller segment lengths equals the largest:      |
| (A, B, C on one line) | AB + BC = AC                                                |
+-----------------------+-------------------------------------------------------------+
| Isosceles Triangle    | Exactly two sides have equal length: AB = AC != BC          |
+-----------------------+-------------------------------------------------------------+
| Equilateral Triangle  | All three sides have equal length: AB = BC = CA             |
+-----------------------+-------------------------------------------------------------+
| Right-Angled Triangle | Satisfies the Pythagorean relation: AB^2 + BC^2 = AC^2      |
+-----------------------+-------------------------------------------------------------+
| Parallelogram         | Opposite sides are equal: AB = CD and BC = DA               |
+-----------------------+-------------------------------------------------------------+
| Rectangle             | Opposite sides are equal AND diagonals are equal:           |
|                       | AB = CD, BC = DA, and AC = BD                               |
+-----------------------+-------------------------------------------------------------+
| Rhombus               | All four sides are equal: AB = BC = CD = DA                 |
+-----------------------+-------------------------------------------------------------+
| Square                | All four sides are equal AND both diagonals are equal:      |
|                       | AB = BC = CD = DA, and AC = BD                              |
+-----------------------+-------------------------------------------------------------+

Equidistant Points and Locus Problems

A frequent problem involves finding a point on a coordinate axis that is equidistant from two given points A and B. If the point lies on the x-axis, its coordinates are P(x, 0). If the point lies on the y-axis, its coordinates are P(0, y). Setting PA^2 = PB^2 eliminates the square root signs and produces an easily solvable linear equation.

Chapter Exercises

Exercise 1: Find the distance between the following pairs of points: (a) (2, 3) and (4, 1); (b) (-5, 7) and (-1, 3); and (c) (a, b) and (-a, -b).

Exercise 2: Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear using the distance formula.

Exercise 3: Find a point on the x-axis which is equidistant from the points A(2, -5) and B(-2, 9).

Solutions and Explanations

Solution 1: For pair (a): d = sqrt((4 - 2)^2 + (1 - 3)^2) = sqrt(2^2 + (-2)^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2) units. For pair (b): d = sqrt((-1 - (-5))^2 + (3 - 7)^2) = sqrt((-1 + 5)^2 + (-4)^2) = sqrt(4^2 + 16) = sqrt(16 + 16) = sqrt(32) = 4sqrt(2) units. For pair (c): d = sqrt((-a - a)^2 + (-b - b)^2) = sqrt((-2a)^2 + (-2b)^2) = sqrt(4a^2 + 4b^2) = sqrt(4(a^2 + b^2)) = 2sqrt(a^2 + b^2) units.

Solution 2: Let the points be A(1, 5), B(2, 3), and C(-2, -11). Compute the three distances: AB = sqrt((2 - 1)^2 + (3 - 5)^2) = sqrt(1^2 + (-2)^2) = sqrt(1 + 4) = sqrt(5). BC = sqrt((-2 - 2)^2 + (-11 - 3)^2) = sqrt((-4)^2 + (-14)^2) = sqrt(16 + 196) = sqrt(212) = 2sqrt(53). AC = sqrt((-2 - 1)^2 + (-11 - 5)^2) = sqrt((-3)^2 + (-16)^2) = sqrt(9 + 256) = sqrt(265). Notice that sqrt(5) is approximately 2.236, 2sqrt(53) is approximately 14.560, and sqrt(265) is approximately 16.278. AB + BC = 2.236 + 14.560 = 16.796 != 16.278. Since the sum of no two segment lengths equals the third length, points A, B, and C are not collinear.

Solution 3: Any point on the x-axis has a y-coordinate of 0. Let the required point be P(x, 0). The point P is equidistant from A(2, -5) and B(-2, 9), which means PA = PB, or PA^2 = PB^2. Compute PA^2: (x - 2)^2 + (0 - (-5))^2 = (x - 2)^2 + 25 = x^2 - 4x + 4 + 25 = x^2 - 4x + 29. Compute PB^2: (x - (-2))^2 + (0 - 9)^2 = (x + 2)^2 + 81 = x^2 + 4x + 4 + 81 = x^2 + 4x + 85. Equate PA^2 = PB^2: x^2 - 4x + 29 = x^2 + 4x + 85. Subtract x^2 from both sides: -4x + 29 = 4x + 85 ==> -8x = 85 - 29 ==> -8x = 56 ==> x = 56 / (-8) = -7. Thus, the required point on the x-axis is (-7, 0).