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Intermediate Geometry : Distance Formula

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Example Questions

Example Question #63 : Lines

Find the length of y = -2x + 3 for the interval $-10 \leq y \leq 15$ .

Possible Answers:

27.951

25.005

22.355

13.463

9.657

Correct answer:

27.951

Explanation:

First, we need to figure out the x-coordinates of the endpoints so that we can use the distance formula, $\sqrt{(x_1 - x_2 )^2 + (y_1 - y_2 ) ^ 2 }$

Plug in -10 for y and solve for x:

-10 = -2x + 3 subtract 3 from both sides

-13 = -2x divide both sides by -2

6.5 = x

Plug in 15 for y and solve for x:

15 = -2x + 3 subtract 3 from both sides

12 = -2x divide both sides by -2

-6 = x

The endpoints are (6.5, -10) and (-6, 15 ) . We could choose either point to be (x_1, y_1) . Let's choose (6.5, -10 ) .

$\sqrt{(6.5 -- 6 )^2 + (-10 - 15 )^2 }= \sqrt { 12.5^2 + (-25)^2 } = \sqrt{156.25 + 625}= \sqrt{781.25} \approx 27.951$

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Example Question #11 : How To Find The Length Of A Line With Distance Formula

Find the length of the line 2x - 4 y + 10 = 0 for the interval $-5 \leq x \leq 0$ .

Possible Answers:

15.811

7.500

4.330

2.500

5.590

Correct answer:

5.590

Explanation:

To calculate the distance, first find the y-coordinates of the endpoints by plugging the x-coordinates into the equation.

First plug in -5

2(-5) - 4y + 10 = 0

-10 -4y + 10 = 0 combining like terms, we get -10 + 10 is 0

-4 y = 0 divide by -4

y = 0

Now plug in 0

2(0 ) - 4 y + 10 = 0

-4y + 10 = 0 subtract 10 from both sides

-4y = -10 divide by -4

y = 2.5

The endpoints are (-5, 0 ) and (0, 2.5) , and now we can plug these points into the distance formula:

$\sqrt{(-5 - 0 )^2 + (0 - 2.5)^2 } = \sqrt{(-5)^2 + (-2.5)^2 } = \sqrt{25 + 6.25} \approx 5.590$

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Example Question #12 : How To Find The Length Of A Line With Distance Formula

Find the length of $y = \frac{1}{4}x + 3$ on the interval $- 8 \leq x \leq 12$ .

Possible Answers:

21.190

6.403

20.616

8.062

8.602

Correct answer:

20.616

Explanation:

To find the length, we need to first find the y-coordinates of the endpoints.

First, plug in -8 for x:

$y = \frac{1}{4} (-8) + 3 = -2 + 3 = 1$

Now plug in 12 for x:

$y = \frac{1}{4} (12 ) + 3 = 3 + 3 = 6$

Our endpoints are (-8, 1) and (12, 6 ) .

To find the length, plug these points into the distance formula:

$\sqrt{(6-1)^2 + (12--8)^2 } = \sqrt{5^2 + 20^2 } = \sqrt{25 + 400 } \approx 20.616$

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Example Question #13 : Distance Formula

Jose is walking from his house to the grocery store. He walks 120 feet north, then turns left to walk another 50 feet west. On the way back home, Jose finds a straight line shortcut back to his house. How long is this shortcut?

Possible Answers:

$150\ ft.$

$170\ ft.$

$90\ ft.$

$130\ ft.$

Correct answer:

$130\ ft.$

Explanation:

When walking north and then taking a left west, a 90 degree angle is formed. When Jose returns home going in a straight line, this will now form the hypotenuse of a right triangle. The legs of the triangle are 120 ft and 50 ft respectively.

To solve, use the pythagorean formula.

$50^{2}+120^{2}=c^{2}$

$2500+14400=c^{2}$

$16900=c^{2}$

$\sqrt{16900}=\sqrt{c^{2}}$

130=c

130 ft is the straight line distance home.

The distance formula could also be used to solve this problem.

We will assume that home is at the point (0,0)

$Distance=\sqrt{(50-0)^{2}+(120-0)^{2}}$

Distance = 130 ft.

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Example Question #61 : Lines

A line has endpoints at (8,4) and (5,10). How long is this line?

Possible Answers:

$\sqrt{5}$

None of these.

$3\sqrt{5}$

$5\sqrt{3}$

Correct answer:

$3\sqrt{5}$

Explanation:

We find the exact length of lines using their endpoints and the distance formula.

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

Given the endpoints,

(x_1,y_1)=(8,4)

(x_2,y_2)=(5, 10)

the distance formula becomes,

$\sqrt{(5-8)^2+(10-4)^2}=\sqrt{9+36}=\sqrt{45}=\sqrt{9\cdot 5}=\mathbf{3\sqrt{5}}$ .

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Example Question #71 : Coordinate Geometry

Find the length of a line with endpoints at (10, 2) and (5, -6) .

Possible Answers:

$\sqrt{91}$

$\sqrt{87}$

$\sqrt{89}$

$\sqrt{85}$

Correct answer:

$\sqrt{89}$

Explanation:

Recall the distance formula for a line with two endpoints $(x_1, y_1)\text{ and }(x_2, y_2)$ :

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

Plug in the given points to find the length of the line:

$\text{distance}=\sqrt{(5-10)^2+(-6-2)^2}=\sqrt{89}$

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Example Question #72 : Coordinate Geometry

A line segment on the coordinate plane has its endpoints at (3.8, -1.7) and (9.2, 5.5) .

Give the length of the segment to the nearest whole tenth.

Possible Answers:

15.2

13.5

14.8

9.0

6.6

Correct answer:

9.0

Explanation:

The distance between endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on the coordinate plane can be calculated using the distance formula

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

Set $x_{1}= 3.8, y_{1} =-1.7,x_{2} =9.2, y_{2}= 5.5$ , and evaluate:

$d = \sqrt{(9.2-3.8)^{2}+[5.5 -(-1.7)]^{2}}$

$= \sqrt{5.4 ^{2}+7.2^{2}}$

$= \sqrt{29.16+51.84}$

$= \sqrt{81}$

= 9 ,

the correct length.

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Intermediate Geometry : Distance Formula

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #63 : Lines

Example Question #11 : How To Find The Length Of A Line With Distance Formula

Example Question #12 : How To Find The Length Of A Line With Distance Formula

Example Question #13 : Distance Formula

Example Question #61 : Lines

Example Question #71 : Coordinate Geometry

Example Question #72 : Coordinate Geometry

All Intermediate Geometry Resources

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