Distance Formula - SSAT Upper Level Quantitative

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A line segment has the endpoints and . What is the length of this line segment?

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Use the following formula to find the distance between two points:

$\text{Distance}$=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$

Plug in the points that are given.

$$\text{Distance}$=$\sqrt{25+25}$=$\sqrt{50}$=$\sqrt{552}$=5\sqrt2$

A line segment has end points at and . What is the length of this line segment?

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The distance between two points is given by the following equation:

$\text{Distance}$=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$

Now, using the two given points, plug them in to find the distance.

Find the length of a line that has end points at $(8, -1)$\text{ and }$(-2, 1)$ .

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Use the distance formula to find the length of this line segment:

$\text{Distance}$=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$

Substitute in the values provided:

At this point, break down the square root:

$\sqrt{104}$=$\sqrt{2\cdot52}$=$\sqrt{2\cdot2\cdot26}$=$\sqrt{2\cdot2\cdot2\cdot13}$

You can remove two of the twos and put them on the outside of the square root symbol, and multiply the two and the thirteen that remain underneath the square root symbol:

$\sqrt{2\cdot2\cdot2\cdot13}$=2$\sqrt{2\cdot13}$=2$\sqrt{26}$

Find the length of the line segment that has endpoints $(0, 4)$\text{ and }$(12, 6)$ .

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and .

Find the length of the line segment that has the endpoints $(2, 4)$\text{ and }$(7, 8)$ .

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Use the distance formula to find the length of the line.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and .

One leg of a triangle has endpoints at the coordinates $(-1, -2)$\text{ and }$(-4, -1)$ . Find the length of this leg.

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Use the distance formula to find the length of the leg.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and .

A line segment has endpoints at $(-2, -1)$\text{ and }$(8, 1)$ . Find the length of this line.

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_1-x_$2)^2$+(y_1-y_$2)^2$}$$

Where,

and .

A side of a square is graphed onto a coordinate plane. The side has endpoints at $(-3, 1)$\text{ and }$(4, -1)$ . Find the length of the side of the square.

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and

The side of a triangle is graphed onto a coordinate plane. The side has endpoints at $(2, 2)$\text{ and }$(1, -5)$ . Find the length of this side.

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and .

Find the length of a line segment that has endpoints at $(1, 2)$\text{ and }$(5, 1)$ .

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and

Find the length of a line segment that has endpoints at $\left($\frac{5}{2}$, $\frac{5}{2}\right)$ and $\left(4, $\frac{4}{5}\right)$ .

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

$(x_1,y_1)=\left($\frac{5}{2}$, $\frac{5}{2}$ \right )$ and $(x_2,y_2)=\left(4,$\frac{4}{5}$ \right )$ .

Find the distance of a line segment that has endpoints at $\left($\frac{1}{2}$, 4\right)$ and $\left($\frac{3}{2}$, 2\right)$ .

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

$(x_1,y_1)=\left($\frac{1}{2}$, 4 \right )$ and $(x_2,y_2)=\left($\frac{3}{2}$, 2 \right )$ .

Find the length of a line segment that has endpoints at $\left($\frac{1}{3}$, 2\right)$ and $\left($\frac{4}{3}$, 3\right)$ .

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

$(x_1,y_1)=\left($\frac{1}{3}$, 2\right)$ and $(x_2,y_2)=\left($\frac{4}{3}$, 3 \right )$ .

Find the length of a line segment that has endpoints at $(2, 1)$\text{ and }$(-1, 0)$ .

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and .

Find the length of a line segment that has endpoints at $(3, 1)$\text{ and }$(-2, -2)$ .

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Use the distance formula to find the length of the line segment.

$D=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Where,

and .

If the two points on a line are $“\left(“$\frac{1}{3}$, $\frac{1}{3}$“\right)”$ and , what is the approximate distance connecting the two points?

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Write the distance formula.

$d=$\sqrt{(y_2-y_$1)^2$+(x_2-x_$1)^2$}$$

Substitute the point values.

$d=$\sqrt{\left( $4-\frac{1}{3}\right)^2$+\left( $2-$\frac{1}{3}\right)^2$} = $\sqrt{\left( $\frac{11}{3}\right)^2$+\left( $$\frac{5}{3}\right)^2$}$

$d=$\sqrt{13.\bar{4}$+2.\bar{7}}= $\sqrt{16.\bar{2}$}= 4.027$

The approximated distance is 4.

Give the length of a segment on the coordinate plane whose endpoints are $\left ($\frac{1}{5}$ , $\frac{4}{5}$ \right )$ and $\left ( -$\frac{3}{5}$, $\frac{1}{5}$ \right )$ .

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The distance between two points and can be obtained using the formula

$d = $\sqrt{\left $(x_{2}$$$-x_{1}$ \right $)^{2}$+\left $(y_{2}$$-y_{1}$ \right $)^{2}$}$

Set :

$d = $\sqrt{\left ( \frac{1}{5}$- \left( - $\frac{3}{5}\right ) \right $)^{2}$+\left ($\frac{4}{5}$-$\frac{1}{5}$ \right $)^{2}$}$

$= $\sqrt{\left( \frac{4}{5}$ \right $)^{2}$+\left ($\frac{3}{5}$ \right $)^{2}$}$

$= $\sqrt{ \frac{16}{25}$ + $\frac{9}{25}$ }$

$= $\sqrt{ \frac{25}{25}$ }$

$= $\sqrt{1}$$

This is not given among the responses.

Give the length of a segment on the coordinate plane whose endpoints are $($\sqrt{11}$, $\sqrt{13}$)$ and the origin.

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The distance between two points and can be obtained using the formula

$d = $\sqrt{\left $(x_{2}$$$-x_{1}$ \right $)^{2}$+\left $(y_{2}$$-y_{1}$ \right $)^{2}$}$

Set :

$d = $\sqrt{\left (\sqrt{11}$-0 \right $)^{2}$+\left( $\sqrt{13}$-0\right $)^{2}$}$

$= $\sqrt{\left (\sqrt{11}$ \right $)^{2}$+\left( $\sqrt{13}\right $)^{2}$}$

$= $\sqrt{11+13}$ = $\sqrt{24}$ = $\sqrt{4}$ \cdot \sqrt {6} = 2 \sqrt {6}$

A segment on the coordinate plane whose endpoints are and the origin has length 10. Calculate .

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The distance between two points and can be obtained using the formula

$d = $\sqrt{\left $(x_{2}$$$-x_{1}$ \right $)^{2}$+\left $(y_{2}$$-y_{1}$ \right $)^{2}$}$

If we set , we can solve for :

$10= $\sqrt{\left (A-0 \right $)^{2}$$+\left (A-0 \right $)^{2}$}$

$10= $\sqrt{ A ^{2}$+A ^{2}}$

$10= $\sqrt{ 2A ^{2}$}$

$100 = $2A^{2 }$$

$100 \div 2 = $2A^{2 }$ \div 2$

$50 = $A^{2}$$

$A = \pm $\sqrt{50 }$ = \pm $\sqrt{25}$ \cdot $\sqrt{2}$ = \pm 5 $\sqrt{2}$$

A segment on the coordinate plane whose endpoints are and the origin has length 10. If is assumed to be positive, calculate .

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The distance between two points and can be obtained using the formula

$d = $\sqrt{\left $(x_{2}$$$-x_{1}$ \right $)^{2}$+\left $(y_{2}$$-y_{1}$ \right $)^{2}$}$

If we set , we can solve for :

$10= $\sqrt{\left (2A-0 \right $)^{2}$$+\left (A-0 \right $)^{2}$}$

$10= $\sqrt{\left(2A \right $)^{2}$$$+A^{2}$}$

$10= $\sqrt{4A ^${2}$+A^{2}$}$

$10= $\sqrt{5A ^{2}$ }$

$100 = $5A^{2}$$

$100\div 5 = $5A^{2}$ \div 5$

$20 = $A^{2}$$

$A = $\sqrt{20}$ = $\sqrt{4}$ \cdot $\sqrt{5}$ = 2$\sqrt{5}$$