Ninth Grade Mathematics: The Distance Formula & Derivation

Measuring the straight-line distance between two points is one of the most fundamental operations in geometry. While distances along horizontal or vertical lines can be found by simple subtraction, computing the diagonal distance between any two arbitrary coordinates requires the synthesis of coordinate geometry and the Pythagorean theorem into the Distance Formula.

Derivation via the Pythagorean Theorem

Let A(x_1, y_1) and B(x_2, y_2) be any two distinct points on the Cartesian coordinate plane. To find the straight-line distance d between them, construct a right-angled triangle by drawing a horizontal line through A and a vertical line through B:

                     y
                     |                     B (x2, y2)
                     |                         +
                     |                        /|
                     |                       / |
                     |                      /  | |y2 - y1|
                     |              d      /   | (Vertical Leg)
                     |                    /    |
                     |                   /     |
                     |                  /      |
                     |       A (x1, y1)+-------+ C (x2, y1)
                     |                 |x2 - x1|
                     |              (Horizontal Leg)
  -------------------+-----------------+-------+---- x
                     |                x1      x2

The horizontal and vertical lines intersect at point C, whose coordinates are (x_2, y_1). Triangle ACB is a right-angled triangle with the right angle at vertex C: The length of the horizontal leg AC is |x_2 - x_1|. The length of the vertical leg BC is |y_2 - y_1|. The hypotenuse AB is the straight-line distance d.

Applying the Pythagorean theorem to triangle ACB:

AB^2 = AC^2 + BC^2
d^2  = (x_2 - x_1)^2 + (y_2 - y_1)^2

Taking the principal (positive) square root of both sides gives the standard Distance Formula:

The Distance Formula:

d = sqrt( (x_2 - x_1)^2 + (y_2 - y_1)^2 )

Because squaring any real number eliminates negative signs—meaning (x_2 - x_1)^2 = (x_1 - x_2)^2—the order in which you subtract the coordinates does not matter. The distance is always symmetric and strictly non-negative.

Distance of a Point from the Origin

A convenient special case occurs when calculating the distance from the origin O(0, 0) to any point P(x, y).

Distance from Origin:
d = sqrt( (x - 0)^2 + (y - 0)^2 ) = sqrt( x^2 + y^2 )

Example: Find the distance from the origin O(0, 0) to point P(-6, 8)
d = sqrt( (-6)^2 + 8^2 ) = sqrt( 36 + 64 ) = sqrt( 100 ) = 10 units.

Step-by-Step Calculation Protocol

To find the distance between any two points:

Find the distance between A(-3, 5) and B(9, -4):

Step 1: Assign coordinates:
        (x_1, y_1) = (-3, 5)  and  (x_2, y_2) = (9, -4)

Step 2: Compute horizontal difference:
        x_2 - x_1 = 9 - (-3) = 9 + 3 = 12

Step 3: Compute vertical difference:
        y_2 - y_1 = -4 - 5 = -9

Step 4: Square both differences:
        (12)^2 = 144
        (-9)^2 = 81

Step 5: Sum the squares and take the square root:
        d = sqrt( 144 + 81 ) = sqrt( 225 ) = 15 units.

Chapter Exercises

Exercise 1: Calculate the exact straight-line distance between the points P(2, -3) and Q(10, 3).

Exercise 2: Find the distance between the points A(-4, -1) and B(2, 7). If the result is a surd, express it both in simplest radical form and as a decimal rounded to two places.

Exercise 3: The distance between the points A(3, y) and B(7, 1) is given as 5 units. Set up an algebraic equation using the distance formula and solve for all possible values of y.

Solutions and Explanations

Solution 1: Identify coordinates: (x_1, y_1) = (2, -3) and (x_2, y_2) = (10, 3). Compute differences: x_2 - x_1 = 10 - 2 = 8, and y_2 - y_1 = 3 - (-3) = 3 + 3 = 6. Apply the distance formula: d = sqrt((8)^2 + (6)^2) = sqrt(64 + 36) = sqrt(100) = 10 units.

Solution 2: Identify coordinates: (x_1, y_1) = (-4, -1) and (x_2, y_2) = (2, 7). Compute differences: x_2 - x_1 = 2 - (-4) = 6, and y_2 - y_1 = 7 - (-1) = 8. Apply the distance formula: d = sqrt((6)^2 + (8)^2) = sqrt(36 + 64) = sqrt(100) = 10 units. The distance is exactly 10 units.

Solution 3: Apply the distance formula with (x_1, y_1) = (3, y) and (x_2, y_2) = (7, 1): d = sqrt((7 - 3)^2 + (1 - y)^2). We are given that d = 5: 5 = sqrt(4^2 + (1 - y)^2) = sqrt(16 + (1 - y)^2). Square both sides of the equation: 25 = 16 + (1 - y)^2. Subtract 16 from both sides: 9 = (1 - y)^2. Take the square root of both sides: 1 - y = 3 or 1 - y = -3. Solving the first case gives y = 1 - 3 = -2. Solving the second case gives y = 1 - (-3) = 4. Thus, there are two possible values for y: y = -2 or y = 4. Both points (3, -2) and (3, 4) are exactly 5 units away from (7, 1).