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Intermediate Geometry : Distance Formula

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Example Questions

Example Question #1 : Distance Formula

A line segment begins from the origin and is 10 units long, which of the following points could NOT be an endpoint for the line segment?

Possible Answers:

(-6,8)

$(7,\sqrt{51})$

(5,9)

$(\sqrt{15},-\sqrt{85})$

(6,8)

Correct answer:

(5,9)

Explanation:

By the distance formula, the sum of the squares of each point must add up to 10 squared. The only point that doesn't fufill this requirement is (5,9)

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Example Question #1 : Distance Formula

A line segment is drawn starting from the origin and terminating at the point (15,8) . What is the length of the line segment?

Possible Answers:

$\sqrt{264}$

$\sqrt{285}$

Correct answer:

Explanation:

Using the distance formula, $\sqrt{8^{2}+15^{2}} = 17$

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Example Question #3 : Distance Formula

What is the distance between (-2,-3) and (6,12) ?

Possible Answers:

Correct answer:

Explanation:

In general, the distance formula is given by: $d=\sqrt{(x_{2}-x_{1})^{2}+{(y_{2}-y_{1})^{2}}}$ and is based on the Pythagorean Theorem.

Let $P_{1}=(-2,-3)$ and $P_{2}= (6,12)$

So the equation to soolve becomes $d=\sqrt{(6-(-2))^{2}+{(12-(3))^{2}}}$ or d=17

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Example Question #3 : Distance Formula

If we graph the equation y=-2x+4 what is the distance from the y-intercept to the x-intercept?

Possible Answers:

$\sqrt{5}$

$2\sqrt{5}$

$2\sqrt{6}$

$\sqrt{6}$

Correct answer:

$2\sqrt{5}$

Explanation:

First, you must figure out where the x and y intercepts lie. To do this we begin by plugging in y=0 to our equation, giving us -2x+4=0 . Thus x=2 . So our x-intercept is the point (2,0) . We then plug in x=0 , giving us y=4 , so we know our y-intercept is the point (0,4) . We then use the distance formula $d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ and plug in our points, giving us $d=\sqrt{(2-0)^2+(0-4)^2}=\sqrt{20}=2\sqrt{5}$

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Example Question #4 : Distance Formula

Find the distance of the line connecting the pair of points

(0,0) and (5, 12) .

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)=(0,0) and (x_2,y_2)=(5,12)

we have

$\sqrt{(5-0)^2+(12-0)^2 }=\sqrt{(5)^2+(12)^2}= \sqrt{25+144} = \sqrt{169} = 13$

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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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Intermediate Geometry : Distance Formula

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #1 : Distance Formula

Example Question #1 : Distance Formula

Example Question #3 : Distance Formula

Example Question #3 : Distance Formula

Example Question #4 : Distance Formula

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

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