Question

A ranger in a lookout tower spots two fires on the campground below. Fire A is 75 meters east
and 40 meters south of the tower. Fire B is 37 meters west and 64 meters south of the tower.
If there is a fire hydrant located three-fourths of the way from Fire A to Fire B, find the location
of the fire hydrant relative to the tower.

Answer 1

Answer:

the fire hydrant is 9 meters west and 58 m south of the tower

Step-by-step explanation:

Fire A and Fire B can be represented as a point in the Cartesian coordinate with the tower as the origin. Fire A is 75 meters east  and 40 meters south of the tower, It can be represented as (75, -40).  Fire B is 37 meters west and 64 meters south of the tower,  It can be represented as (-37, -64).

If a point O(x, y) divides a line segment AB in the ratio n:m, with A($x_1,y_1$) and B($x_2,y_2$), the point O(x, y) is at:

$x=\frac{n}{n+m}(x_2-x_1)+x_1\\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1$

The fire hydrant (x, y) divides Fire A(75, -40) and fire B(-37, -64) in the ratio 3:1(three fourth), hence:

$x=\frac{n}{n+m}(x_2-x_1)+x_1=\frac{3}{4}(-37-75) +75=\frac{3}{4}(-112)+75=-84+75\\x=-9 \\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1=\frac{3}{4}(-64-(-40))+(-40)= \frac{3}{4}(-18)-40=-18-40\\y=-58$

The fire hydrant is at (-9, -58). That is the fire hydrant is 9 meters west and 58 m south of the tower

Answer 2

The location of the hydrant relative to the tower is $\left(-9, -58 \right)$ meters.

How to locate the coordinates of the hydrant relative to the tower

In this question we should use the vectorial formula for a line segment to determine the coordinates of the hydrant relative to the tower ($\vec C$), in meters:

$\vec C = \vec A + r\cdot (\vec B - \vec A)$ (1)

Where:

• $\vec A$ - Coordinates of the fire A relative to the tower, in meters.
• $\vec B$ - Coordinates of the fire B relative to the tower, in meters.
• $r$ - Distance ratio.

Now we determine the location of the hydrant: ($\vec A = (75, -40)\,[m]$, $\vec B = (-37, -64)\,[m]$, $r = \frac{3}{4}$)

$\vec C = (75,-40) + \frac{3}{4}\cdot [(-37,-64)-(75,-40)]$

$\vec C = (75,-40) + \frac{3}{4} \cdot (-112,-24)$

$\vec C = (75,-40) + \left(-84, -18\right)$

$\vec C = \left(-9, -58 \right)\,[m]$

The location of the hydrant relative to the tower is $\left(-9, -58 \right)$ meters. $\blacksquare$

To learn more on vectors, we kindly invite to check this verified question: https://brainly.com/question/13188123