Question

Point R and S are plotted on the coordinate plane below.



R is (-8, -6)

S is (4, 2)



What is the distance between points R and S, to the nearest tenth of a unit?


20.0


7.2


6.3


14.4

Answer 1

To find the distance between the points R(-8, -6) and S(4, 2) on the coordinate plane, we can use the distance formula:

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

where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of points R and S.

Let’s assign:
- \(R(x_1, y_1) = (-8, -6)\)
- \(S(x_2, y_2) = (4, 2)\)

Now plug in the values:

\[
d = \sqrt{(4 - (-8))^2 + (2 - (-6))^2}
\]
\[
d = \sqrt{(4 + 8)^2 + (2 + 6)^2}
\]
\[
d = \sqrt{(12)^2 + (8)^2}
\]
\[
d = \sqrt{144 + 64}
\]
\[
d = \sqrt{208}
\]

Now, we can simplify \(\sqrt{208}\):

\[
\sqrt{208} = \sqrt{16 \times 13} = 4\sqrt{13}
\]

Now we need to calculate \(4\sqrt{13}\):

\(\sqrt{13} \approx 3.60555\)

Thus,

\[
4\sqrt{13} \approx 4 \times 3.60555 \approx 14.4222
\]

Rounding to the nearest tenth gives us:

\[
d \approx 14.4
\]

So, the distance between points R and S, to the nearest tenth of a unit, is **14.4**.