In the standard (x, y) coordinate plane, what is the distance, in coordinate units, betwee… — ACT

In the standard $(x, y)$ coordinate plane, what is the distance, in coordinate units, between the points $(1, 2)$ and $(4, 6)$ ?

The distance is

5

coordinate units.

The correct answer is

5

because applying the distance formula

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

to the points

(1, 2)

and

(4, 6)

yields

\sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5

Identify the coordinates of the two given points.

Let

(x_1, y_1) = (1, 2)

and

(x_2, y_2) = (4, 6)

This establishes the coordinate values for the distance formula.

Calculate the difference between the

x

-coordinates and the difference between the

y

-coordinates.

x_2 - x_1 = 4 - 1 = 3

and

y_2 - y_1 = 6 - 2 = 4

These differences represent the horizontal and vertical side lengths of the right triangle formed by the two points.

Apply the distance formula

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

d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Squaring the coordinate differences, adding them, and taking the square root yields the straight-line distance.

Using the distance formula to find the straight-line distance between two points in the coordinate plane.

Question