How to find the length of a line with distance formula - SSAT Upper Level Quantitative

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Question

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

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Answer

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

$\text{Distance}=\sqrt{(5-1)^2+(1-2)^2}=\sqrt{16+1}=\sqrt{17}$

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Question

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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Answer

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 $(x_2,y_2)=\left(\frac{4}{3}, 3 \right )$ .

$\text{Distance}=\sqrt{(\frac{4}{3}-\frac{1}{3})^2+(3-2)^2}=\sqrt{1+1}=\sqrt2$

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Question

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

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Answer

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 .

$\text{Distance}=\sqrt{(-1-2)^2+(0-1)^2}=\sqrt{9+1}=\sqrt{10}$

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Question

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

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Answer

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 .

$\text{Distance}=\sqrt{(-2-3)^2+(-2-1)^2}=\sqrt{25+9}=\sqrt{34}$

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Question

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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Answer

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

$\text{Distance}=\sqrt{(4-\frac{5}{2})^2+(\frac{4}{5}-\frac{5}{2})^2}=\sqrt{\frac{9}{4}+\frac{289}{100}}=2.27$

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Question

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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Answer

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 $(x_2,y_2)=\left(\frac{3}{2}, 2 \right )$ .

$\text{Distance}=\sqrt{(\frac{3}{2}-\frac{1}{2})^2+(2-4)^2}=\sqrt{1+4}=\sqrt{5}$

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Question

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

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Answer

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.

$\text{Distance}=\sqrt{(4-(-1))^2+(-2-3)^2}$

$\text{Distance}=\sqrt{(5)^2+(-5)^2}$

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

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Question

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

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Answer

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.

$\text{Distance}=\sqrt{(-2-8)^2+(-3-1)^2}=\sqrt{100+16}=\sqrt{116}=10.77$

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Question

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

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Answer

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:

$\text{Distance}=\sqrt{(-2-8)^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}$

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:

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

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Question

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

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Answer

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 .

$\text{Distance}=\sqrt{(12-0)^2+(6-4)^2}=\sqrt{144+4}=\sqrt{148}=12.166$

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Question

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

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Answer

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 .

$\text{Distance}=\sqrt{(7-2)^2+(8-4)^2}=\sqrt{25+16}=\sqrt{41}$

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Question

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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Answer

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 .

$\text{Length}=\sqrt{(-4-(-1))^2+(-1-(-2))^2}=\sqrt{9+1}=\sqrt{10}$

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Question

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

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Answer

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 .

$\text{Distance}=\sqrt{(8-(-2))^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}=2\sqrt{26}$

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Question

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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Answer

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

$\text{Distance}=\sqrt{(4-(-3))^2+(-1-1)^2}=\sqrt{49+4}=\sqrt{53}$

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Question

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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Answer

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 .

$\text{Distance}=\sqrt{(1-2)^2+(-5-2)^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt2$

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Question

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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Answer

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.

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Question

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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Answer

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Set $x_{1}= \frac{1}{5}, y_{1}= \frac{4}{5}, x_{2}= -\frac{3}{5}, y_{2} = \frac{1}{5}$ :

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

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

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

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

$= \sqrt{1}$

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Question

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

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Answer

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Set $x_{1}= y_{1}=0, x_{2}= \sqrt{11}, y_{2} = \sqrt{13}$ :

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

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

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

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Question

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

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Answer

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ 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 $x_{1}= y_{1} = 0, x_{2}= y_{2} = A, d=10$ , we can solve for :

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

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

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

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

$100 = 2A^{2 }$

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

$50 = A^{2}$

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

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Question

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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Answer

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ 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 $x_{1}= y_{1} = 0, x_{2}=2A, y_{2} = A, d=10$ , we can solve for :

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

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

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

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

$10^{2}=\left ( \sqrt{ 5A ^{2}} \right )^{2}$

$100 = 5A^{2}$

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

$20 = A^{2}$

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

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