Question

question.

A straight, solid line connects plotted points upper A and upper B on a coordinate plane. The line slopes downward from left to right. Point upper A is left parenthesis negative 3 comma 5 right parenthesis. Point upper B is left parenthesis 7 comma 2 right parenthesis.

To the nearest tenth of a unit, what is the distance between points A and B?

(1 point)
Responses

9.0 units
9.0 units

10.2 units
10.2 units

10.4 units
10.4 units

10.0 units

Answer 1

To find the distance between points A and B, we can use the distance formula:

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

Here, the coordinates of point A are \((-3, 5)\) and the coordinates of point B are \((7, 2)\).

We will assign:
- \((x_1, y_1) = (-3, 5)\)
- \((x_2, y_2) = (7, 2)\)

Now we can calculate the differences:

\[
x_2 - x_1 = 7 - (-3) = 7 + 3 = 10
\]
\[
y_2 - y_1 = 2 - 5 = -3
\]

Now substitute these values into the distance formula:

\[
d = \sqrt{(10)^2 + (-3)^2}
\]
\[
d = \sqrt{100 + 9}
\]
\[
d = \sqrt{109}
\]

Now we can approximate \(\sqrt{109}\):

Using a calculator:

\(\sqrt{109} \approx 10.4403\)

To the nearest tenth of a unit, the distance is:

\[
10.4 \text{ units}
\]

Therefore, the correct response is:

**10.4 units**