SAT II Math I : Distance Formula

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #171 : Geometry

Find the distance between \(\displaystyle (3, 4)\) and \(\displaystyle (-12, -5)\).

Possible Answers:

\(\displaystyle 3 \sqrt {34}\)

\(\displaystyle \sqrt {82}\)

\(\displaystyle 18\)

\(\displaystyle 10 \sqrt {3}\)

\(\displaystyle 2 \sqrt {6}\)

Correct answer:

\(\displaystyle 3 \sqrt {34}\)

Explanation:

For this problem we will need to use the distance formula:

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

In our case,

\(\displaystyle (x_1,y_1)=(-12, -5)\) and \(\displaystyle (x_2,y_2)=(3, 4)\).

Plugging these values into the formula we are able to find the distance.

\(\displaystyle d = \sqrt {(3-(-12))^2 + (4-(-5)))^2))}\)

\(\displaystyle =\sqrt {15^2+9^2}\)

\(\displaystyle = \sqrt {306}\)

\(\displaystyle = \sqrt {9 \cdot 34} = 3 \sqrt {34}\)

Example Question #171 : Geometry

What is the distance between \(\displaystyle (1,6)\) and \(\displaystyle (-3,-2)\)?

Possible Answers:

\(\displaystyle 4\sqrt5\)

\(\displaystyle 2\sqrt5\)

\(\displaystyle \frac{4}{3}\)

\(\displaystyle 5\sqrt4\)

\(\displaystyle 4\sqrt2\)

Correct answer:

\(\displaystyle 4\sqrt5\)

Explanation:

Write the distance formula.

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

Substitute the points:  

\(\displaystyle (x_1,y_1)=(1,6)\)

\(\displaystyle (x_2,y_2)=(-3,-2)\)

\(\displaystyle d=\sqrt{(-3-1)^2+(-2-6)^2}\)

\(\displaystyle d=\sqrt{16+64} = \sqrt{80}\)

The radical can be broken down into factors of perfect squares.

\(\displaystyle \sqrt{80} = \sqrt{16}\cdot \sqrt5 = 4\sqrt5\)

The answer is:  \(\displaystyle 4\sqrt5\)

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