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High School Math : How to find the length of a line with distance formula

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Example Questions

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Example Question #11 : Midpoint And Distance Formulas

What is the length of a line segment with end points (0,1) and (5,7) ?

Possible Answers:

9.13

$\sqrt{61}$

$\sqrt{68}$

7.25

Correct answer:

$\sqrt{61}$

Explanation:

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in the given values and solve for the length.

(5-0)^2+(7-1)^2=l^2

(5)^2+(6)^2=l^2

25+36=l^2

61=l^2

$\sqrt{61}=\sqrt{l^2}$

$\sqrt{61}=l$

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Example Question #2 : How To Find The Length Of A Line With Distance Formula

What would be the length of a line with endpoints at (1,6) and (9,9) ?

Possible Answers:

8.13

$\sqrt{73}$

$\sqrt{39}$

$\sqrt{91}$

4.5

Correct answer:

$\sqrt{73}$

Explanation:

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in the given values and solve for the length.

(9-1)^2+(9-6)^2=l^2

(8)^2+(3)^2=l^2

64+9=l^2

73=l^2

$\sqrt{73}=\sqrt{l^2}$

$\sqrt{73}=l$

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Example Question #1361 : High School Math

What line goes through the points (-1,5) and (5,15) ?

Possible Answers:

-4x-3y=9

-2x+5y=10

-5x+3y=20

3x+2y=5

3x-5y=15

Correct answer:

-5x+3y=20

Explanation:

Find the slope between the two points:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{15-5}{5-(-1)}=\frac{10}{6}=\frac{5}{3}$

Next, use the slope-intercept form of the equation:

y=mx+b or $15=\frac{5}{3}(5)+b$ where $b=\frac{20}{3}$

So the equation becomes $y=\frac{5}{3}x+ \frac{20}{3}$ or in standard form -5x+3y=20 .

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Example Question #1362 : High School Math

What is the length of a line with endpoints at (4,10) and (8,13) ?

Possible Answers:

625

Correct answer:

Explanation:

The formula for the length of a line is very similiar to the pythagorean theorem:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in our given numbers to solve:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

(8-4)^2+(13-10)^2=l^2

4^2+3^2=l^2

16+9=l^2

25=l^2

$\sqrt{25}=\sqrt{l^2}$

5=l

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Example Question #1363 : High School Math

What is the length of a line with endpoints at (6,13) and (9,17) ?

Possible Answers:

$\frac{4}{3}$

625

$\frac{3}{4}$

Correct answer:

Explanation:

The formula for the length of a line is very similiar to the pythagorean theorem:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in our given numbers to solve:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

(17-13)^2+(9-6)^2=l^2

4^2+3^2=l^2

16+9=l^2

25=l^2

$\sqrt{25}=\sqrt{l^2}$

5=l

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Example Question #1364 : High School Math

Find the distance between points (2,-4) and (14,-2) .

Possible Answers:

$\sqrt{144}$

144

148

$\sqrt{148}$

Correct answer:

$\sqrt{148}$

Explanation:

Use the distance formula:

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

Substitute the given points into the formula:

$d=\sqrt{(14-2)^2+(-2-(-4))^2}$

$d=\sqrt{148}$

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Example Question #1365 : High School Math

What is the length of a line with endpoints of (2,-4) and (7,9) ?

Possible Answers:

3.7

17.1

14 7

13.9

Correct answer:

13.9

Explanation:

The distance formula is just a reworking of the Pythagorean theorem: $l=\sqrt{\Delta x^2+\Delta y^2}$

Expand that.

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

Plug in our given values.

$l=\sqrt{(7-2)^2+(9--4)^2}$

$l=\sqrt{(5)^2+(13)^2}$

$l=\sqrt{25+169}$

$l=\sqrt{194}$

l=13.9

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Example Question #1366 : High School Math

What is the distance between (1,5) and (6,17) ?

Possible Answers:

Correct answer:

Explanation:

Let $P_{1}=(1,5)$ and $P_{2} = (6,17)$ .

The distance formula is given by $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Substitute in the given points:

$d = \sqrt{(6-1)^{2}+(17-5)^{2}}=\sqrt{(5)^{2}+(12)^{2}}=\sqrt{25+144}=\sqrt{169}=13$

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Example Question #1367 : High School Math

A line segment has endpoints at (2,3) and (-7,3) . What is the distance of this segment?

Possible Answers:

6.36

4.5

Correct answer:

Explanation:

To find the distance, we use the distance formula: $d^2=\Delta x^2+\Delta y^2$ .

Expand that:

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

Plug in our given values.

d^2=(-7-2)^2+(3-3)^2

d^2=(-9)^2+(0)^2

d^2=81+0

$\sqrt{d^2}=\sqrt{81}$

d=9

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

If a line has a length of , and the endpoints are (0,0) and (4,y) , what is the value of ?

Possible Answers:

1.5

Correct answer:

Explanation:

The formula for the length of a line, l, is the distance formula, which is very similar to the Pythagorean Theorem.

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Note that the problem has already given us a value for the length of the line. That means l=5 . Plug in all of the given values and solve for the missing term.

(4-0)^2+(y-0)^2=5^2

4^2+y^2=5^2

16+y^2=25

Subtract $\small 16$ from both sides.

16+y^2-16=25-16

y^2=9

$\sqrt{y^2}=\sqrt{9}$

y=3

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High School Math : How to find the length of a line with distance formula

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 : Midpoint And Distance Formulas

Example Question #2 : How To Find The Length Of A Line With Distance Formula

Example Question #1361 : High School Math

Example Question #1362 : High School Math

Example Question #1363 : High School Math

Example Question #1364 : High School Math

Example Question #1365 : High School Math

Example Question #1366 : High School Math

Example Question #1367 : High School Math

Example Question #1 : How To Find The Length Of A Line With Distance Formula

All High School Math Resources

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