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Intermediate Geometry : Lines

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Example Questions

Example Question #5 : Distance Formula

Find the distance of the line connecting the pair of points

(1, 2) and (4, 6) .

Possible Answers:

Correct answer:

Explanation:

By the distance formula

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

where (x_1,y_1)=(1,2) and (x_2,y_2)=(4,6)

we have

$\sqrt{(4-1)^2+(6-2)^2 }=\sqrt{(3)^2+(4)^2}= \sqrt{9+16} = \sqrt{25} = 5$

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Example Question #3 : How To Find The Length Of A Line With Distance Formula

Find the length of $\small y = \frac{1}{5} x - 2$ for $\small -2 \leq x \leq 5$

Possible Answers:

$\small \sqrt{36.56}$

$\small \sqrt{26.96}$

$\small \small \sqrt{60.56}$

$\small \small \sqrt{50.96}$

$\small \sqrt{9.36}$

Correct answer:

$\small \small \sqrt{50.96}$

Explanation:

To find the distance, first we have to find the specific coordinate pairs that we're finding the distance between. We know the x-values, so to find the y-values we can plug these endpoint x-values into the line:

$\small y = \frac{1}{5}(-2)-2$ first multiply

$\small y = -0.4 - 2$ then subtract

$\small y = -2.4$

$\small y = \frac{1}{5}(5)-2$ first multiply

$\small y = 1 - 2$ then subtract

$\small y = -1$

Now we know that we're finding the distance between the points $\small \small (5, -1)$ and $\small (-2, -2.4)$ . We can just plug these values into the distance formula, using the first pair as $\small (x_{1}, y_{1})$ and the second pair as $\small (x_{2}, y_{2})$ . It would work either way since we are squaring these values, this just makes it easier.

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

$\small \sqrt{(5--2)^2 + (-1 - -2.4)^2 }$

$\small \sqrt{(5+2)^2 + (-1 + 2.4)^2}$

$\small \sqrt{7^2 + 1.4^2}$

$\small \sqrt{49+1.96} = \sqrt{50.96}$

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Example Question #3 : Distance Formula

Find the length of $\small y = -3x + 4$ for $\small 0 \leq x \leq 5$

Possible Answers:

$4\sqrt{17}$

$\small \sqrt{554}$

$5\sqrt{10}$

$\small \small \sqrt{74}$

Correct answer:

$5\sqrt{10}$

Explanation:

$\small \small y = -3(0)+4$ first multiply

$\small \small y = 0 +4$ then add

$\small \small y = 4$

$\small \small y =-3(5)+4$ first multiply

$\small \small y = -15 +4$ then add

$\small \small y = -11$

Now we know that we're finding the distance between the points $\small (0,4)$ and $\small (5, -11)$ . We can just plug these values into the distance formula, using the first pair as $\small (x_{1}, y_{1})$ and the second pair as $\small (x_{2}, y_{2})$ . Note that it would work either way since we are squaring these values anyway.

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

$\small \small \sqrt{(0-5)^2 + (4 - -11)^2 }$

$\small \small \sqrt{(-5)^2 + (15)^2}$

$\small \small \sqrt{25 + 225} = \sqrt{250}$

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Example Question #4 : How To Find The Length Of A Line With Distance Formula

Find the length of the line $\small y= \frac{1}{2}x - 6$ for $\small -1 \leq y \leq 8$

Possible Answers:

$\small \sqrt{733}$

$9\sqrt{5}$

$\small \sqrt{481}$

$\small \sqrt{137.25}$

$\small \sqrt{373}$

Correct answer:

$9\sqrt{5}$

Explanation:

To find the distance, first we have to find the specific coordinate pairs that we're finding the distance between. We know the y-values, so to find the x-values we can plug these endpoint y-values into the line:

$\small -1 = \frac{1}{2}x - 6$ add 6 to both sides

$\small 5 = \frac{1}{2} x$ multiply by 2

10 = x this endpoint is (10, -1)

$\small 8 = \frac{1}{2} x -6$ add 6 to both sides

$\small 14 = \frac{1}{2}x$ multiply by 2

$\small 28 = x$ this endpoint is (28, 8)

Now we can plug these two endpoints into the distance formula:

$\small \sqrt{(x_{1}- x_{2})^2 + (y_{1} - y_{2})^2 }$ note that it really does not matter which pair we use as $\small (x_{1}, y_{1})$ and which as $\small (x_{2}, y_{2})$ since we'll be squaring these differences anyway, just as long as we are consistent.

$\small \sqrt{(28 - 10)^2 + (8 - - 1 )^2 }$

$\small \sqrt{18^2 + 9^2 }$

$\small \sqrt{324 + 81 } = \sqrt{405}=9\sqrt{5}$

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Example Question #6 : Distance Formula

Find the length of $y = \frac{1}{5} x + 7$ for the interval $-5 \leq x \leq 10$ .

Possible Answers:

15.297

5.099

5.831

16.155

15.033

Correct answer:

15.297

Explanation:

To find this length, we need to know the y-coordinates for the endpoints.

First, plug in -5 for x:

$y = \frac{1}{5}(-5) + 7 = -1 + 7 = 6$

Next, plug in 10 for x:

$y = \frac{1}{5} (10 ) + 7 = 2 + 7 = 9$

So we are finding the distance between the points (-5, 6) and (10, 9 )

We will use the distance formula, $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2 ) ^2 }$ . We could assign either point as (x_1, y_1) and it would still work, but let's choose (10, 9 ) :

$\sqrt{(10 - - 5 )^2 + (9 - 6 )^ 2 } = \sqrt{15^2 + 3^2 } = \sqrt{225 + 9 } = \sqrt{234} \approx 15.297$

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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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Intermediate Geometry : Lines

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #5 : Distance Formula

Example Question #3 : How To Find The Length Of A Line With Distance Formula

Example Question #3 : Distance Formula

Example Question #4 : How To Find The Length Of A Line With Distance Formula

Example Question #6 : Distance Formula

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

All Intermediate Geometry Resources

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