Geometry - How to find the length of a line with distance formula

Find the distance of the line connecting the pair of points

and .

Explanation

By the distance formula

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

where and

we have

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

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

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 and .

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

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

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

$5\sqrt{10}$

$4\sqrt{17}$

$\small \small \sqrt{74}$

$\small \sqrt{554}$

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:

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

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

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

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

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

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?

$130\ ft.$

$170\ ft.$

$150\ ft.$

$90\ ft.$

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 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.

Find the length of a line with endpoints at and .

$\sqrt{89}$

$\sqrt{91}$

$\sqrt{87}$

$\sqrt{85}$

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:

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

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

$\small \small \sqrt{50.96}$

$\small \small \sqrt{60.56}$

$\small \sqrt{26.96}$

$\small \sqrt{36.56}$

$\small \sqrt{9.36}$

Explanation

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

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

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

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

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

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

Find the distance of the line connecting the pair of points

and .

Explanation

By the distance formula

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

where and

we have

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

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

$9\sqrt{5}$

$\small \sqrt{373}$

$\small \sqrt{733}$

$\small \sqrt{481}$

$\small \sqrt{137.25}$

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

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.

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

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

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

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

$3\sqrt{5}$

$\sqrt{5}$

$5\sqrt{3}$

None of these.

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,

the distance formula becomes,

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

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

$2\sqrt{5}$

$\sqrt{5}$

$\sqrt{6}$

$2\sqrt{6}$

Explanation

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 $d=\sqrt{(2-0)^2+(0-4)^2}=\sqrt{20}=2\sqrt{5}$

How to find the length of a line with distance formula

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