Question

Each unit in the coordinate plane corresponds to 1 mile. Find the distance from the school to Cherry Street. Round your answer to the nearest tenth.

Answer 1

The distance from the school to Cherry Street is approximately 11.5 miles (rounded to the nearest tenth).

To find the distance from the school (9, 6) to Cherry Street, we can use the distance formula:

`\[ d = √((x_2 - x_1)^2 + (y_2 - y_1)^2) \]`

The school is at (9, 6), and Cherry Street is on the line
`\( y = -(1)/(2)x - 2 \)`. We need to find the point of intersection between the line and the line passing through the school.

Substitute x = 9 into the equation of Cherry Street:

`\[ y = -(1)/(2)(9) - 2 \]`

`\[ y = -(11)/(2) \]`

So, the point of intersection is (9, -
`\((11)/(2)\)`).

Now, apply the distance formula:

`\[ d = \sqrt{(9 - 9)^2 + \left(\left(-(11)/(2)\right) - 6\right)^2} \]`

`\[ d = \sqrt{0 + \left(-(23)/(2)\right)^2} \]`

`\[ d = \sqrt{(529)/(4)} \]`

`\[ d = (√(529))/(2) \]`

`\[ d = (23)/(2) \]`

Therefore, the distance from the school to Cherry Street is approximately 11.5 miles (rounded to the nearest tenth).

Answer 2

`11.2`

Explanation:

We can use the distance formula to solve this problem.

This distance formula states that the distance between two points
`(x_1, y_1)` and
`(x_2,y_2)` is equal to
`√((\Delta x)^2+(\Delta y)^2)=√((x_2-x_1)^2+(y_2-y_1)^2)`.

School is at coordinate point (9,6) and the closest point to school on Cherry St. is (4,-4).

Thus, the distance between school and Cherry Street is
`√((9-4)^2+(6-(-4))^2)=√(5^2+10^2)=√(125)=5√(5)\approx \boxed{11.2}`