Distance Formula

SAT Subject Test in Math II · Learn by Concept

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SAT Subject Test in Math II › Distance Formula

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1

What is the distance between the point (1,2) and (8,5)?

$\sqrt58$

CORRECT

0

0

0

Explanation

For this question we will use the distance formula to solve.

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

In our case

and

Substituting these values in we get the following

$d=\sqrt((8-1)^2+(5-2)^2)$

$d=\sqrt(7^2+3^2)$

$d=\sqrt(49+9)$

$d=\sqrt(58)$

2

The points A=(-2,0), B=(0,3), and C=(0,0) makes a triangle. What is the distance between point A and point B?

$\sqrt13$

CORRECT

$\sqrt15$

0

0

0

Explanation

For this question we need to use the distance formula for points A and B.

Point A will be our and point B will be

Now we substitute these values into the following:

$d=\sqrt((0--2)^2+(3-0)^2)$

$d=\sqrt(2^2+3^2)$

$d=\sqrt(4+9)$

$d=\sqrt13$

3

Find the distance between the two points (2,7) and (4,6).

$\sqrt{5}$

CORRECT

$\sqrt{3}$

0

$\sqrt{205}$

0

$\sqrt{17}$

0

$\sqrt{3}$

0

Explanation

The distance between two points is found using the formula $d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

For this problem the values are as follows:

$x_{1}=2$

$x_{2}=4$

$y_{1}=7$

$y_{2}=6$

Input the values into the formula and simplify

$d=\sqrt{(4-2)^2+(7-6)^2}=\sqrt{(2)^2+(1)^2}=\sqrt{4+1}=\sqrt{5}$

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