Distance Formula

SAT Subject Test in Math I · Learn by Concept

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SAT Subject Test in Math I › Distance Formula

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1

What is the distance between and ?

$4\sqrt5$

CORRECT

$\frac{4}{3}$

0

$2\sqrt5$

0

$4\sqrt2$

0

$5\sqrt4$

0

Explanation

Write the distance formula.

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

Substitute the points:

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

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

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

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

The answer is: $4\sqrt5$

2

Find the distance between and .

$3 \sqrt {34}$

CORRECT

0

$2 \sqrt {6}$

0

$\sqrt {82}$

0

$10 \sqrt {3}$

0

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,

and .

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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