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ACT Math : Distance Formula

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Example Questions

Example Question #1 : Distance Formula

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√2

4√2

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 gives 7√2

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Example Question #2 : 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

9.3

12.5

11.4

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

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Example Question #1 : Distance Formula

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

Possible Answers:

$\dpi{100} \small 5.0$

$\dpi{100} \small 1.0$

$\dpi{100} \small 7.8$

$\dpi{100} \small 3.2$

$\dpi{100} \small 6.4$

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 $\sqrt{(-2-2)^{2}+(-3-2)^{2}}=\sqrt{(-4)^{2}+(-5)^{2}}=\sqrt{16+25}=\sqrt{41}\approx 6.4$ .

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Example Question #2 : Distance Formula

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

Possible Answers:

$5$

$\frac{5}{2}$

$-4$

$\frac{5}{4}$

$\frac{-5}{4}$

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 $\frac{-3-2}{-2-2}=\frac{-5}{-4}=\frac{5}{4}$ .

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Example Question #3 : Distance Formula

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

Possible Answers:

Correct answer:

Explanation:

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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Example Question #4 : 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

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

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Example Question #5 : Distance Formula

What is the distance between points (3,7) and (-1,-1) ?

Possible Answers:

$\sqrt{80}$

$\sqrt{12}$

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:

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

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

$16+64=c^{2}$

$80=c^{2}$

$\sqrt{80}=c$

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Example Question #6 : 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:

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

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Example Question #6 : Distance Formula

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

Possible Answers:

100

Correct answer:

Explanation:

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

Actmath_29_372_q6_4_copy

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

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Example Question #3 : Distance Formula

Line segment has end points of (1,5) and (3,-3) .

Line segemet has end points of (10,12) and (4,14) .

What is the distance between the midpoints?

Possible Answers:

Correct answer:

Explanation:

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

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

$AB\ mid = (2,1)$ 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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ACT Math : Distance Formula

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : Distance Formula

Example Question #2 : Distance Formula

Example Question #1 : Distance Formula

Example Question #2 : Distance Formula

Example Question #3 : Distance Formula

Example Question #4 : Distance Formula

Example Question #5 : Distance Formula

Example Question #6 : Distance Formula

Example Question #6 : Distance Formula

Example Question #3 : Distance Formula

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