Question

Beth Anne marks the points (6, 1) and (-9, 6) on the coordinate plane. What is the distance between the two points, rounded to the nearest tenth of a unit?

Answer 1

To find the distance between the points \((6, 1)\) and \((-9, 6)\) in the coordinate plane, we use the distance formula, which is given by:

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

Here, we can set \((x_1, y_1) = (6, 1)\) and \((x_2, y_2) = (-9, 6)\).

Now we can substitute into the formula:

1. Calculate \(x_2 - x_1\):

\[
x_2 - x_1 = -9 - 6 = -15
\]

2. Calculate \(y_2 - y_1\):

\[
y_2 - y_1 = 6 - 1 = 5
\]

3. Substitute these values back into the distance formula:

\[
d = \sqrt{(-15)^2 + 5^2}
\]

Calculating each term:

\[
(-15)^2 = 225
\]
\[
5^2 = 25
\]

So, we have:

\[
d = \sqrt{225 + 25} = \sqrt{250}
\]

Now, we can simplify \(\sqrt{250}\):

\[
\sqrt{250} = \sqrt{25 \times 10} = \sqrt{25} \times \sqrt{10} = 5\sqrt{10}
\]

Next, we will approximate \(\sqrt{10}\). Since \(\sqrt{10} \approx 3.162\), we can calculate:

\[
5\sqrt{10} \approx 5 \times 3.162 \approx 15.81
\]

Finally, rounding \(15.81\) to the nearest tenth, we find:

\[
\text{Distance} \approx 15.8
\]

Thus, the distance between the points \((6, 1)\) and \((-9, 6)\) is \(\boxed{15.8}\).