Distance Formula - Intermediate Geometry

Card 0 of 17

Question

A line segment is drawn starting from the origin and terminating at the point . What is the length of the line segment?

Answer

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

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Question

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?

Answer

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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Question

What is the distance between and ?

Answer

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

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Question

If we graph the equation what is the distance from the y-intercept to the x-intercept?

Answer

First, you must figure out where the x and y intercepts lie. To do this we begin by plugging in to our equation, giving us . Thus . So our x-intercept is the point . We then plug in , giving us , so we know our y-intercept is the point . 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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Question

Find the distance of the line connecting the pair of points

and .

Answer

By the distance formula

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

where and

we have

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

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Question

Find the distance of the line connecting the pair of points

and .

Answer

By the distance formula

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

where and

we have

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

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Question

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

Answer

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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Question

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

Answer

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:

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

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

Answer

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

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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Question

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

Answer

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 and

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

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

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Question

Find the length of for the interval $-10 \leq y \leq 15$ .

Answer

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:

subtract 3 from both sides

divide both sides by -2

Plug in 15 for y and solve for x:

subtract 3 from both sides

divide both sides by -2

The endpoints are and . We could choose either point to be . Let's choose .

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

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Question

Find the length of the line for the interval $-5 \leq x \leq 0$ .

Answer

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

First plug in -5

combining like terms, we get -10 + 10 is 0

divide by -4

Now plug in 0

subtract 10 from both sides

divide by -4

The endpoints are and , 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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Question

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

Answer

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 and .

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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Question

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?

Answer

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 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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Question

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

Answer

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,

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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Question

Find the length of a line with endpoints at and .

Answer

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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Question

A line segment on the coordinate plane has its endpoints at and .

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

Answer

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:

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

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

$= \sqrt{29.16+51.84}$

$= \sqrt{81}$

,

the correct length.

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