In the standard (x, y) coordinate plane, what is the distance between the points (1, \frac… — ACT

In the standard $(x, y)$ coordinate plane, what is the distance between the points $(1, \frac{1}{2})$ and $(5, \frac{7}{2})$ ?

The distance between the two points is

5

The distance

d

between

(x_1, y_1)

and

(x_2, y_2)

is given by the formula

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

. Substituting

(1, \frac{1}{2})

and

(5, \frac{7}{2})

gives

d = \sqrt{(5-1)^2 + (\frac{7}{2} - \frac{1}{2})^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5

Calculate the horizontal change (difference in

x

-coordinates)

x_2 - x_1 = 5 - 1 = 4

This determines the horizontal leg of the right triangle formed between the two points.

Calculate the vertical change (difference in

y

-coordinates)

y_2 - y_1 = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3

This determines the vertical leg of the right triangle formed between the two points.

Apply the Pythagorean theorem (distance formula) to find the distance

d

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

The straight-line distance is the hypotenuse of the right triangle with legs of lengths 4 and 3.

The Distance Formula

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