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

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Example Question #111 : Lines

Let W and Z be the points of intersection between the parabola whose graph is y = –x² – 2x + 3, and the line whose equation is y = x – 7. What is the length of the line segment WZ?

 

Possible Answers:

7

4√2

7√2

4

Correct answer:

7√2

Explanation:

First, set the two equations equal to one another. 

x² – 2x + 3 = x – 7

 

Rearranging gives

x² + 3x – 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 Actmath_7_113_q1gives 7√2 

 

Example Question #1 : How To Find The Length Of A Line With Distance Formula

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

Possible Answers:

7.5

12.5

11.4

9.3

Correct answer:

11.4

Explanation:

Use the distance formula:

D = √((y2 – y1)2 + (x2 – x1)2)

D = √((6 + 3)2 + (5 + 2)2)

D = √((9)2 + (7)2)

D = √(81 + 49)

D = √130

D = 11.4

Example Question #744 : Act Math

Coordinates

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

Possible Answers:

\dpi{100} \small 7.8

\dpi{100} \small 6.4

\dpi{100} \small 5.0

\dpi{100} \small 1.0

\dpi{100} \small 3.2

Correct answer:

\dpi{100} \small 6.4

Explanation:

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.

Example Question #2 : How To Find The Length Of A Line With Distance Formula

Coordinates

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

Possible Answers:

5

\frac{-5}{4}

\frac{5}{2}

\frac{5}{4}

-4

Correct answer:

\frac{5}{4}

Explanation:

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}.

Example Question #3 : Distance Formula

What is the distance between  and ?

Possible Answers:

Correct answer:

Explanation:

Let  and  and use the distance formula: .  The distance formula is a specific application of the more general Pythagorean Theorem:  a^{2} + b^{2} = c^{2}.

Example Question #1 : Distance Formula

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

Possible Answers:

\sqrt{113}

15

113

\sqrt{7}

\sqrt{15}

Correct answer:

\sqrt{113}

Explanation:

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}

Example Question #6 : How To Find The Length Of A Line With Distance Formula

What is the distance between points  and ?

Possible Answers:

\sqrt{12}

\sqrt{80}

Correct answer:

\sqrt{80}

Explanation:

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

Example Question #2 : Distance Formula

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

Possible Answers:

Correct answer:

Explanation:

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

 

Example Question #3 : Distance Formula

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

Possible Answers:

100

5

20

25

10

Correct answer:

10

Explanation:

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

Actmath_29_372_q6_1

Actmath_29_372_q6_2

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

Actmath_29_372_q6_3

Actmath_29_372_q6_4_copy

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

 

Example Question #4 : Distance Formula

Line segment  has end points of  and .

Line segemet  has end points of  and .

What is the distance between the midpoints?

Possible Answers:

Correct answer:

Explanation:

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})

and 

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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