On a coordinate map of a harbor, a lighthouse is located at L(-4, 9) and a dock is located… — ACT

On a coordinate map of a harbor, a lighthouse is located at $L(-4, 9)$ and a dock is located at $D(8, -7)$ . A buoy is positioned at the midpoint of the straight-line segment connecting the lighthouse and the dock. If a boat is anchored at $B(-1, 5)$ , what is the distance, in coordinate units, between the boat and the buoy?

Answer: 5 coordinate units

Answer

The distance between the boat and the buoy is 5 units.

To find the distance between the boat and the buoy, first determine the location of the buoy. Since the buoy is at the midpoint of the segment connecting the lighthouse at

L(-4, 9)

and the dock at

D(8, -7)

, we use the midpoint formula:

M = \left(\frac{-4 + 8}{2}, \frac{9 + (-7)}{2}\right) = (2, 1)

. Next, find the distance between the boat at

B(-1, 5)

and the buoy at

M(2, 1)

using the distance formula:

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

Step-by-Step Solution

Find the coordinates of the buoy, which is the midpoint

M(x_m, y_m)

of the segment connecting the lighthouse

L(-4, 9)

and the dock

D(8, -7)

M(2, 1)

The midpoint formula is given by

x_m = \frac{x_1 + x_2}{2}

and

y_m = \frac{y_1 + y_2}{2}

. Substituting the coordinates of

L

and

D

, we get

x_m = \frac{-4 + 8}{2} = 2

and

y_m = \frac{9 + (-7)}{2} = 1

Calculate the distance

d

between the boat at

B(-1, 5)

and the buoy at

M(2, 1)

5

units

The distance formula is given by

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

. Substituting the coordinates of

B

and

M

yields

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

Key Concept

Applying the midpoint and distance formulas sequentially to solve a coordinate geometry word problem.

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