Distance Formula - SAT Math

Card 0 of 20

Question

Find the distance between and .

Answer

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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Question

What is the distance between and ?

Answer

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: $4\sqrt5$

Compare your answer with the correct one above

Question

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

Answer

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

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

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

$d=\sqrt(49+9)$

$d=\sqrt(58)$

Compare your answer with the correct one above

Question

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?

Answer

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:

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

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

$d=\sqrt(4+9)$

$d=\sqrt13$

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

Find the distance between and .

Answer

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.

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

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

$= \sqrt {306}$

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

Compare your answer with the correct one above

Question

What is the distance between and ?

Answer

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: $4\sqrt5$

Compare your answer with the correct one above

Question

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

Answer

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

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

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

$d=\sqrt(49+9)$

$d=\sqrt(58)$

Compare your answer with the correct one above

Question

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?

Answer

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:

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

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

$d=\sqrt(4+9)$

$d=\sqrt13$

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

What is the distance between and ?

Answer

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: $4\sqrt5$

Compare your answer with the correct one above

Question

Find the distance between and .

Answer

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.

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

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

$= \sqrt {306}$

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

$d=\sqrt(49+9)$

$d=\sqrt(58)$

Compare your answer with the correct one above

Question

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?

Answer

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:

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

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

$d=\sqrt(4+9)$

$d=\sqrt13$

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

What is the distance between and ?

Answer

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: $4\sqrt5$

Compare your answer with the correct one above

Question

Find the distance between and .

Answer

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.

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

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

$= \sqrt {306}$

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

$d=\sqrt(49+9)$

$d=\sqrt(58)$

Compare your answer with the correct one above

Question

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?

Answer

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:

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

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

$d=\sqrt(4+9)$

$d=\sqrt13$

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

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