Find the distance between $(-2, 1)$ and $(4, 9)$.

How we read your problem

Find the straight-line (Euclidean) distance between two points in the plane.

Step-by-step solution

1. 1

   Label the points

   $$(x_1, y_1) = (-2, 1),\quad (x_2, y_2) = (4, 9)$$
2. 2

   Find the horizontal and vertical differences

   $$\Delta x = 6,\quad \Delta y = 8$$
3. 3

   Square the differences and add them

   $$(6)^2 + (8)^2 = 36 + 64 = 100$$
4. 4

   Take the square root

   $$d = \sqrt{100} = 10$$
5. 5

   Midpoint (average of the coordinates)

   $$M = \left(\frac{-2 + 4}{2}, \frac{1 + 9}{2}\right) = (1, 5)$$

Formulas used

• Distance formula

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

• Midpoint formula

  $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

The distance formula is the Pythagorean theorem applied to the right triangle whose legs are the horizontal and vertical differences between the points.

Verification

• d² = 100 matches Δx² + Δy² = 100