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

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Question

Let W and Z be the points of intersection between the parabola whose graph is y = –x_² – 2_x + 3, and the line whose equation is y = x – 7. What is the length of the line segment WZ?

Answer

First, set the two equations equal to one another.

–x_² – 2_x + 3 = x – 7

Rearranging gives

x_² + 3_x – 10 = 0

Factoring gives

(x + 5)(x – 2) = 0

The points of intersection are therefore W(–5, –12) and Z(2, –5)

Using the distance formula gives 7√2

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Question

What is the area of a square with a diagonal that has endpoints at (4, –1) and (2, –5)?

Answer

First, we need to find the length of the diagonal. In order to do that, we will use the distance formula:

Now that we have the length of the diagonal, we can find the length of the side of a square. The diagonal of a square makes a 45/45/90 right triangle with the sides of the square, which we shall call s. Remember that all sides of a square are equal in length.

Because this is a 45/45/90, the length of the hypotenuse is equal to the length of the side multiplied by the square root of 2

The area of the square is equal to s2, which is 10.

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Question

In an _xy-_plane, what is the length of a line connecting points at (–2,–3) and (5,6)?

Answer

Use the distance formula:

D = √((_y_2 – _y_1)2 + (_x_2 – _x_1)2)

D = √((6 + 3)2 + (5 + 2)2)

D = √((9)2 + (7)2)

D = √(81 + 49)

D = √130

D = 11.4

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Question

What is the distance between points $\dpi{100} \small A$ and $\dpi{100} \small B$ , to the nearest tenth?

Answer

The distance between points $\dpi{100} \small A$ and $\dpi{100} \small B$ is 6.4. Point $\dpi{100} \small A$ is at $\dpi{100} \small (-2,-3)$ . Point $\dpi{100} \small B$ is at $\dpi{100} \small (2,2)$ . Putting these points into the distance formula, we have $\sqrt{(-2-2)^{2}+(-3-2)^{2}}=\sqrt{(-4)^{2}+(-5)^{2}}=\sqrt{16+25}=\sqrt{41}\approx 6.4$ .

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Question

What is the slope of the line between points $\dpi{100} \small A$ and $\dpi{100} \small B$ ?

Answer

The slope of the line between points $\dpi{100} \small A$ and $\dpi{100} \small B$ is $\frac{5}{4}$ . Point $\dpi{100} \small A$ is at $\dpi{100} \small (-2,-3)$ . Point $\dpi{100} \small B$ is at $\dpi{100} \small (2,2)$ . Putting these points into the slope formula, we have $\frac{-3-2}{-2-2}=\frac{-5}{-4}=\frac{5}{4}$ .

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Question

In the standard coordinate plane, what is the perimeter of a triangle with vertices at , and .

Answer

This problem is a combination of two mathematical principles: the distance formula between two points, and the reduction of radicals.

To begin this problem we must find the lengths of all sides of the triangle. Because we have the coordinates of the end points of each side, we can apply the distance formula.

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ = distance between the two points $\dpi{100} (x_{1},y_{1})$ and $(x_{2},y_{2})$ .

The application of this formula by writing out all the symbols and inserting the points and perhaps using a calculator is cumbersome and can be time consuming. There is a faster application of this formula in a geometric sense. Follow the steps below to give this process a try. This should be repeated for each distance you are trying to find and in our case, it is three different distances.

Draw a right triangle with one leg horizontal and one vertical.

For the horizontal leg, find the distance between the coordinates of the two choosen points and write down that distance: $(x_{2}-x_{1})$ .

Repeat for the vertical leg substituting the coordinates: $(y_{2}-y_{1})$ .

Apply the Pythagorean Theorem, where and are the horizontal and vertical leg in no particular order.

Solve for .

is the length of one of the sides of your triangle in which you wish to find the perimeter. If the formula is faster for you, use it. If the geometric method is faster and easier to visualize, use that one instead.

After applying the formula to all three legs, you’ll find that you have the lengths, $\sqrt{52}$ , $\sqrt{52}$ , and . Once you add them together to find the perimeter (the perimeter of a triangle is all sides added together), you have the value $\dpi{100} 8+2\sqrt{52}$ . Unfortunately you won’t find the answer as you still need to simplify the radical.

To go about reducing the radical you will need to break down 52 into its prime factors: 13, 2, and 2. (As an aside, any even number can be checked for prime factors by dividing it by two until it is no longer even.) Since there are two 2’s under the radical, we can rewrite the radical as $\sqrt{2\cdot 2\cdot 13}$ or $\dpi{100} \sqrt{2\cdot 2} \cdot \sqrt{13}$ . Since $\sqrt{4}=2$ we can reduce $\sqrt{52}$ to $2\sqrt{13}$ .

Therefore, $\dpi{100} 8+2\sqrt{52}$ can be rewritten $8+2(2)\sqrt{13}$ or $8+4\sqrt{13}$ , which is our perimeter.

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Question

What is the distance between $\left ( 1,-2 \right )$ and $\left ( 6,10 \right )$ ?

Answer

Let $P_{1}=(1,-2)$ and $P_{2}=(6,10)$ and use the distance formula: $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ . The distance formula is a specific application of the more general Pythagorean Theorem: $a^{2} + b^{2} = c^{2}$ .

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Question

What is the distance, in coordinate units, between the points $(-2,6)$ and $(5,-2)$ in the standard $(x,y)$ coordinate plane?

Answer

The distance formula is $\sqrt{((x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2})}=d$ , where $d$ = distance.

Plugging in our values, we get

$d=\sqrt{((5-(-2))^{2}+(6-(-2))^{2}}=\sqrt{7^{2}+8^{2}}=\sqrt{49+64}=\sqrt{113}$

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Question

What is the distance between points and ?

Answer

Solution A:

Use the distance formula to calculate the distance between the two points:

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

$d=\sqrt{(-1-3)^{2}+(-1-7)^{2}}$

$d=\sqrt{(-4)^{2}+(-8)^{2}}$

$d=\sqrt{16+64}$

$d=\sqrt{80}$

Solution B:

Draw the two points on a coordinate graph and create a right triangle with sides 4 and 5. Using the Pythagorean Theorem, solve for the hypotenuse or the distance between the two points:

$a^{2}+b^{2}=c^{2}$

$4^{2}+8^{2}=c^{2}$

$16+64=c^{2}$

$80=c^{2}$

$\sqrt{80}=c$

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Question

What is the distance between (1,5) and (6,17)?

Answer

Let $P_{1}=(1,5)$ and $P_{2}=(6,17)$

So we use the distance formula $d =\sqrt{(x_{2} - x_{1})^2+(y_{2} - y_{1})^2}$

and evaluate it using the given points:

$d=\sqrt{(6-1)^2+(17 - 5)^2}= \sqrt{(5)^2+(12)^2}=13$

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Question

Line segment has end points of and .

Line segemet has end points of and .

What is the distance between the midpoints?

Answer

The midpopint is found by taking the average of each coordinate:

$P_{mid} = (\frac{x_{1}+x_{2} }{2},\frac{y_{1}+y_{2} }{2})$

$AB\ mid = (2,1)$ and $CD\ mid = \left ( 7,13 \right )$

The distance formula is given by

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

Making the appropriate substitutions we get a distance of 13.

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Question

What is the distance of the line

Between and ?

Round to the nearest hundredth.

Answer

What is the distance of the line

Between and ?

To calculate this, you need to know the points for these two values. To find these, substitute in for the two values of given:

Likewise, do the same for :

Now, this means that you have two points:

and

The distance formulat between two points is:

$\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$

For our data, this is:

$\sqrt{(7-4)^2+(10-12)^2}=\sqrt{(3)^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}$

This is:

or approximately

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Question

What is the distance between the x and y intercepts of:

Round to the nearest hundredth.

Answer

In order to find the distance between these two points, we first need the points! To find them, substitute in for x and y. Thus, for each you get:

Thus, the two points are:

and

The distance formulat between two points is:

$\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$

For your data, this is very simply:

$\sqrt{(12)^2+(24)^2}=\sqrt{576 + 144} = \sqrt{720}$

$\sqrt{720}=26.83281572999748....$ or

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Question

Two towns—Town A and Town B—are represented by points on a map overlaid with a standard x- and y-coordinate plane. Town A and Town B are represented by points and , respectively. If each unit on the map represents an actual distance of 20 miles, which of the following is closest to the actual distance, in miles, between these two towns?

Answer

In order to find the actual distance between these two towns, we must find the distance between these two points. We can find the distance between the points by using the distance formula, where the variable, , represents the distance:

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

The variables $x_{2}$ and $x_{1}$ represent the x-values at each point and $y_{2}$ and $y_{1}$ represent the y-values at each point. We can calculate the distance by substituting the x- and y-coordinates of each point into the distance formula.

$d = \sqrt{(-2-3)^2+ (15-3)^2}$

$d= \sqrt{169}$

Since each unit on the map represents an actual distance of miles, the actual distance between the two cities can be calculated using the the following operation:

$13 \times20 $ miles = 260 miles.$

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Question

What is the length of the line segment whose endpoints are:

Answer

To find the distance between two points, use the distance formula:

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

$d = \sqrt{(4 - 0)^2 + (1 - -2)^2}$

$d = \sqrt{(4 )^2 + (3)^2}$

$d = \sqrt{16+9}$

$d=\sqrt{25}$

this gives .

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Question

In an _xy-_plane, what is the length of a line connecting points at (–2,–3) and (5,6)?

Answer

Use the distance formula:

D = √((_y_2 – _y_1)2 + (_x_2 – _x_1)2)

D = √((6 + 3)2 + (5 + 2)2)

D = √((9)2 + (7)2)

D = √(81 + 49)

D = √130

D = 11.4

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Question

Let W and Z be the points of intersection between the parabola whose graph is y = –x_² – 2_x + 3, and the line whose equation is y = x – 7. What is the length of the line segment WZ?

Answer

First, set the two equations equal to one another.

–x_² – 2_x + 3 = x – 7

Rearranging gives

x_² + 3_x – 10 = 0

Factoring gives

(x + 5)(x – 2) = 0

The points of intersection are therefore W(–5, –12) and Z(2, –5)

Using the distance formula gives 7√2

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Question

What is the area of a square with a diagonal that has endpoints at (4, –1) and (2, –5)?

Answer

First, we need to find the length of the diagonal. In order to do that, we will use the distance formula:

Now that we have the length of the diagonal, we can find the length of the side of a square. The diagonal of a square makes a 45/45/90 right triangle with the sides of the square, which we shall call s. Remember that all sides of a square are equal in length.

Because this is a 45/45/90, the length of the hypotenuse is equal to the length of the side multiplied by the square root of 2

The area of the square is equal to s2, which is 10.

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Question

What is the distance between points $\dpi{100} \small A$ and $\dpi{100} \small B$ , to the nearest tenth?

Answer

The distance between points $\dpi{100} \small A$ and $\dpi{100} \small B$ is 6.4. Point $\dpi{100} \small A$ is at $\dpi{100} \small (-2,-3)$ . Point $\dpi{100} \small B$ is at $\dpi{100} \small (2,2)$ . Putting these points into the distance formula, we have $\sqrt{(-2-2)^{2}+(-3-2)^{2}}=\sqrt{(-4)^{2}+(-5)^{2}}=\sqrt{16+25}=\sqrt{41}\approx 6.4$ .

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Question

What is the slope of the line between points $\dpi{100} \small A$ and $\dpi{100} \small B$ ?

Answer

The slope of the line between points $\dpi{100} \small A$ and $\dpi{100} \small B$ is $\frac{5}{4}$ . Point $\dpi{100} \small A$ is at $\dpi{100} \small (-2,-3)$ . Point $\dpi{100} \small B$ is at $\dpi{100} \small (2,2)$ . Putting these points into the slope formula, we have $\frac{-3-2}{-2-2}=\frac{-5}{-4}=\frac{5}{4}$ .

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