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SAT II Math I : Distance Formula

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Example Questions

Example Question #1 : Distance Formula

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

Possible Answers:

$3 \sqrt {34}$

$2 \sqrt {6}$

$10 \sqrt {3}$

$\sqrt {82}$

Correct answer:

$3 \sqrt {34}$

Explanation:

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

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

In our case,

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

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

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

$=\sqrt {15^2+9^2}$

$= \sqrt {306}$

$= \sqrt {9 \cdot 34} = 3 \sqrt {34}$

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Example Question #2 : Distance Formula

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

Possible Answers:

$4\sqrt2$

$2\sqrt5$

$4\sqrt5$

$5\sqrt4$

$\frac{4}{3}$

Correct answer:

$4\sqrt5$

Explanation:

Write the distance formula.

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

Substitute the points:

(x_1,y_1)=(1,6)

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

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

$d=\sqrt{16+64} = \sqrt{80}$

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

$\sqrt{80} = \sqrt{16}\cdot \sqrt5 = 4\sqrt5$

The answer is: $4\sqrt5$

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SAT II Math I : Distance Formula

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

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Example Question #1 : Distance Formula

Example Question #2 : Distance Formula

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