Hello, and welcome to today's mathematics lesson. Today, we are going to explore the Distance Formula. By the end of this lesson, you will understand how to find the distance between any two points on a coordinate plane, how to apply this formula to solve various problems, and how to use it to identify special properties of geometric figures.
Let us begin with the foundation. Imagine you have two points plotted on a graph. How would you find out exactly how far apart they are? You might think of using a ruler, but in coordinate geometry, we have something far more powerful — a formula that works every single time.
Consider two points, A and B. Let point A have coordinates (x₁, y₁) and point B have coordinates (x₂, y₂). Now, picture a right-angled triangle formed by drawing horizontal and vertical lines from these points. The horizontal side of this triangle represents the difference between the x-coordinates, which is (x₂ – x₁). The vertical side represents the difference between the y-coordinates, which is (y₂ – y₁).
Here comes the beautiful connection. The line joining our two points, A and B, forms the hypotenuse of this right-angled triangle. And we know from Pythagoras' theorem that the square of the hypotenuse equals the sum of the squares of the other two sides.
Therefore, the distance between points A and B is given by the distance formula: The distance equals √[(x₂ – x₁)² + (y₂ – y₁)²]. This can also be written as √[(x₁ – x₂)² + (y₁ – y₂)²], since squaring removes the sign difference.
Notice something important here. Since squaring removes any negative sign, it does not matter whether you write (x₂ – x₁) or (x₁ – x₂) — the result is identical. The same applies to the y-coordinates. This formula works for points in any quadrant of the coordinate plane, as long as you use the correct signs for each coordinate.
Let us see this formula in action. Suppose we want to find the distance between the points (3, 6) and (0, 2). We identify x₁ = 3, y₁ = 6, x₂ = 0, and y₂ = 2. The difference in x-coordinates is 0 – 3 = –3, and squaring this gives 9. The difference in y-coordinates is 2 – 6 = –4, and squaring this gives 16. Adding these, we get 25, and the square root of 25 is 5. So the distance between these points is exactly 5 units.
Here is a special case that often appears in problems. What is the distance of any point from the origin? The origin has coordinates (0, 0), so if our point is (x, y), the distance formula simplifies beautifully. The distance from the origin equals √(x² + y²).
For example, the distance from the origin to the point (–12, 5) is √(144 + 25), which equals √169, giving us 13 units. Similarly, for the point (15, –8), we get √(225 + 64), which equals √289, giving 17 units.
Now, let us think about points that lie on the axes. Any point on the x-axis has a y-coordinate of zero, so we write it as (x, 0). Any point on the y-axis has an x-coordinate of zero, so we write it as (0, y). This knowledge helps us solve many interesting problems.
Suppose we need to find points on the x-axis that are 5 units away from (6, –3). We let our unknown point be (x, 0) and apply the distance formula. This gives us the equation: 5 equals √[(x – 6)² + (0 + 3)²]. Squaring both sides, we get 25 equals x² – 12x + 36 + 9. Simplifying, x² – 12x + 20 = 0. Solving this quadratic equation yields x equals 2 or x equals 10. Therefore, the required points are (2, 0) and (10, 0).
Here is another elegant application. We can use the distance formula to prove that three points are collinear — that is, they lie on the same straight line. Three points A, B, and C are collinear if the sum of two of the distances equals the third distance. This means either AB plus BC equals AC, or AB plus AC equals BC, or AC plus BC equals AB.
Let us verify this with points A (1, –1), B (6, 4), and C (4, 2). First, we calculate AB: the difference in x is 5, the difference in y is 5, so √(25 + 25) equals 5√2. Next, BC: the difference in x is 2, the difference in y is 2, so √(4 + 4) equals 2√2. And AC: the difference in x is 3, the difference in y is 3, so √(9 + 9) equals 3√2. Notice that 2√2 + 3√2 = 5√2, which equals AB. Since BC plus AC equals AB, the points are indeed collinear, with C lying between A and B.
The distance formula also helps us identify special triangles. Consider points A (8, 3), B (0, 9), and C (14, 11). Calculating all three sides: AB involves differences of 8 and 6, giving √(64 + 36) equals 10. BC involves differences of 14 and 2, giving √(196 + 4) equals 10√2. And CA involves differences of 6 and 8, giving √(36 + 64) equals 10. Since AB equals CA, the triangle is isosceles. Furthermore, checking Pythagoras' theorem: AB squared plus CA squared equals 100 plus 100, which is 200, and this equals BC squared. Therefore, the triangle is not just isosceles but also right-angled.
We can extend this idea to quadrilaterals. A quadrilateral is a parallelogram if both pairs of opposite sides are equal. It is a rectangle if opposite sides are equal and the diagonals are also equal. It is a rhombus if all four sides are equal. And it is a square if all sides are equal and the diagonals are equal as well. The distance formula gives us the tool to verify all these properties by calculation.
Let us explore one more powerful application: finding the circumcentre of a triangle. The circumcentre is the point equidistant from all three vertices of the triangle. If you draw a circle with this point as centre and any vertex on the circumference, all three vertices will lie on that circle.
To find the circumcentre of a triangle with vertices A (4, 6), B (0, 4), and C (6, 2), we let the circumcentre be P with coordinates (x, y). Since PA equals PB, we set up an equation using the distance formula. Squaring both sides: x² – 8x + 16 + y² – 12y + 36 equals x² + y² – 8y + 16. Simplifying, we obtain 2x + y = 9. Let us call this equation one.
Similarly, since PA equals PC, we set up another equation. Squaring and simplifying similarly: x² – 8x + 16 + y² – 12y + 36 equals x² – 12x + 36 + y² – 4y + 4. This gives x – 2y = –3. Let us call this equation two. Solving equations one and two simultaneously, we find x = 3 and y = 3. Therefore, the circumcentre is at (3, 3).
Before we conclude, let me walk you through a practical problem involving circles. Suppose we need to find the area of a circle with centre at (5, –3) that passes through (–7, 2). The radius is simply the distance between these two points. Calculating: the differences are 12 and 5, so √(144 + 25) equals √169, which is 13. The area equals πr², which is 3.14 × 169, giving us 530.66 square units.
Let us recap the key takeaways from today's lesson.
First, the distance between two points (x₁, y₁) and (x₂, y₂) is given by √[(x₂ – x₁)² + (y₂ – y₁)²].
Second, the distance from any point (x, y) to the origin is √(x² + y²).
Third, points on the x-axis have the form (x, 0) and points on the y-axis have the form (0, y).
Fourth, three points are collinear if the sum of two distances equals the third.
Fifth, we can classify triangles and quadrilaterals by comparing distances between their vertices.
And sixth, the circumcentre of a triangle is found by solving for the point equidistant from all three vertices.
The distance formula is one of the most elegant results in coordinate geometry. It connects algebra and geometry through the simple yet profound Pythagorean relationship. Master this formula, and you hold the key to solving countless problems involving position, length, and shape.
Keep practicing, stay curious, and remember — every point in the plane has a story to tell through its coordinates. Until next time, happy learning!