Question

find the distance between the pair of points.
e(7, -4), f(-2,8)

d = \square (simplify your answer. type an exact answer, using radicals as needed.)

Explanation:

Step1: Recall distance formula

The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
For points \( E(7, -4) \) and \( F(-2, 8) \), let \( x_1 = 7 \), \( y_1 = -4 \), \( x_2 = -2 \), \( y_2 = 8 \).

Step2: Substitute values into formula

Calculate \( x_2 - x_1 = -2 - 7 = -9 \) and \( y_2 - y_1 = 8 - (-4) = 12 \).
Then, \( d = \sqrt{(-9)^2 + 12^2} \).

Step3: Simplify the expression

\( (-9)^2 = 81 \) and \( 12^2 = 144 \), so \( d = \sqrt{81 + 144} = \sqrt{225} = 15 \)? Wait, no, \( 81 + 144 = 225 \)? Wait, \( 81 + 144 = 225 \), but \( \sqrt{225}=15 \)? Wait, no, wait: \( (-9)^2 = 81 \), \( 12^2 = 144 \), sum is \( 81 + 144 = 225 \), square root of 225 is 15? Wait, but let's check again. Wait, \( 9^2 = 81 \), \( 12^2 = 144 \), 81 + 144 = 225, square root of 225 is 15. Wait, but is that correct? Wait, the x difference is -9, y difference is 12. So squared terms are 81 and 144, sum 225, square root 15. Wait, but let me recalculate: \( 7 - (-2) = 9 \)? Wait no, the formula is \( (x_2 - x_1) \), so \( x_2 = -2 \), \( x_1 = 7 \), so \( -2 - 7 = -9 \), squared is 81. \( y_2 = 8 \), \( y_1 = -4 \), so \( 8 - (-4) = 12 \), squared is 144. Sum is 81 + 144 = 225, square root is 15. Wait, but that seems too straightforward. Wait, maybe I made a mistake? Wait, no, 9-12-15 triangle is a Pythagorean triple. So yes, the distance is 15. Wait, but let's do it again. The distance formula is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). So \( x_2 - x_1 = -2 - 7 = -9 \), \( y_2 - y_1 = 8 - (-4) = 12 \). Then square both: \( (-9)^2 = 81 \), \( 12^2 = 144 \). Sum: 81 + 144 = 225. Square root of 225 is 15. So the distance is 15.

Answer:

\( \boxed{15} \)