Distance between two points on a plane

The Distance Between Two Points on a Plane

Consider two points on a coordinate plane:

1. Point A with coordinates $A(x_1, y_1)$
2. Point B with coordinates $B(x_2, y_2)$

To find the distance between points A and B, we first draw a straight line segment joining the two points. Next, we draw dotted lines from each point to form a right-angled triangle, with AB as the hypotenuse.

From the figure (you can insert an appropriate diagram here), you will see that:

• AC = $|x_2 - x_1|$ (horizontal distance)
• BC = $|y_2 - y_1|$ (vertical distance)

Since triangle ABC is a right-angled triangle, we apply the Pythagorean Theorem:

$AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Taking the square root of both sides:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Example 1

Find the distance between points $A(3, 4)$ and $B(10, 12)$.

$x_1 = 3, \, y_1 = 4, \, x_2 = 10, \, y_2 = 12$

Distance = $\sqrt{(10 - 3)^2 + (12 - 4)^2}$

= $\sqrt{7^2 + 8^2}$

= $\sqrt{49 + 64}$

= $\sqrt{113} \approx$ 10.63 units

Example 2

Find the distance between points $A(5, 2)$ and $B(-5, -10)$.

$x_1 = 5, \, y_1 = 2, \, x_2 = -5, \, y_2 = -10$

Distance = $\sqrt{(-5 - 5)^2 + (-10 - 2)^2}$

= $\sqrt{(-10)^2 + (-12)^2}$

= $\sqrt{100 + 144}$

= $\sqrt{244} \approx$ 15.62 units

Special Case

If $AC = 5$ and $BC = 12$, then:

Distance $AB = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} =$ 13 units.