A triangle has vertices on a coordinate grid at R, left parenthesis, minus, 3, comma, 3, right parenthesisR(−3,3), S, left parenthesis, minus, 3, comma, minus, 9, right parenthesisS(−3,−9), and T, left parenthesis, minus, 9, comma, minus, 9, right parenthesis, .T(−9,−9). What is the length, in units, of start overline, R, S, end overline

Question

RS
?

Answer 1

To find the length of the segment \( \overline{RS} \), we will use the distance formula between the points \( R(-3, 3) \) and \( S(-3, -9) \).

The distance formula is:

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

In our case, \( R(x_1, y_1) = (-3, 3) \) and \( S(x_2, y_2) = (-3, -9) \).

Substituting the coordinates into the distance formula, we have:

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

Calculating the differences:

• The \( x \)-coordinates: \( -3 - (-3) = 0 \)
• The \( y \)-coordinates: \( -9 - 3 = -12 \)

Now substituting back into the formula:

\[ d = \sqrt{(0)^2 + (-12)^2} \]

Calculating the squares:

\[ d = \sqrt{0 + 144} = \sqrt{144} = 12 \]

Thus, the length of \( \overline{RS} \) is \( \boxed{12} \) units.