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SAT Math : Lines

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Example Question #19 : Distance Formula

In terms of , give the length of a segment on the coordinate plane with endpoints $\left (- \frac{1}{4} , N \right )$ and $\left (N, - \frac{1}{4} \right )$ .

Possible Answers:

$\sqrt{2N^{2}+ N}$

$\sqrt{2N^{2}- N}$

$\sqrt{2N^{2}+ \frac{1}{8}}$

$\sqrt{2N^{2}+N+\frac{1}{8}}$

$\sqrt{2N^{2}-N+\frac{1}{8}}$

Correct answer:

$\sqrt{2N^{2}+N+\frac{1}{8}}$

Explanation:

The length of a segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be calculated using the distance formula:

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

Setting $x_{1} = y_{2} = -\frac{1}{4}$ and $x_{2} = y_{1} = N$ , and substituting:

$d = \sqrt{ \left (N - \left (- \frac{1}{4} \right )\right ) ^{2} + \left ( - \frac{1}{4} - N \right ) ^{2}}$

$d = \sqrt{ \left (N + \frac{1}{4} \right ) ^{2} + \left ( - N - \frac{1}{4} \right ) ^{2}}$

The binomials can be rewritten using the perfect square trinomial pattern:

$d = \sqrt{ \left [N^{2}+ 2 \cdot N \cdot \frac{1}{4} + \left ( \frac{1}{4}\right ) ^{2} \right ] + \left [(-N)^{2} - 2 \cdot (-N) \cdot\left ( - \frac{1}{4}\right ) + \left ( - \frac{1}{4}\right ) ^{2} \right ] }$

$d = \sqrt{ \left ( N^{2}+ \frac{1}{2} N + \frac{1}{16} \right ) +\left ( N^{2}+ \frac{1}{2} N + \frac{1}{16} \right ) }$

Simplify and collect like terms:

$d = \sqrt{ N^{2}+ N^{2}+ \frac{1}{2} N + \frac{1}{2} N + \frac{1}{16} + \frac{1}{16} }$

$= \sqrt{2 N^{2}+ N + \frac{1}{8} }$

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Example Question #20 : Distance Formula

Give the length, in terms of , of a line segment on the coordinate plane whose endpoints are (N, N) and $\left ( \frac{1}{3}, -\frac{1}{3} \right )$ .

Possible Answers:

$\sqrt{2N^{2}+\frac{4}{3}N+\frac{2}{9}}$

$\sqrt{2N^{2}+\frac{4}{3}N }$

$\sqrt{2N^{2}+\frac{2}{9}}$

$\sqrt{2N^{2}-\frac{4}{3}N }$

$\sqrt{2N^{2}-\frac{4}{3}N+\frac{2}{9}}$

Correct answer:

$\sqrt{2N^{2}+\frac{2}{9}}$

Explanation:

The length of a segment with endpoints $(x_{1},y_{1}) , (x_{2},y_{2})$ can be calculated using the distance formula

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

Substituting $x_{1} = \frac{1}{3} ,y_{1} =-\frac{1}{3}, x_{2} = y_{2}= N$ , and simplifying"

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

$= \sqrt{\left (N - \frac{1}{3}\right )^{2}+ \left (N - \left ( - \frac{1}{3} \right ) \right )^{2} }$

$= \sqrt{\left (N - \frac{1}{3}\right )^{2}+ \left (N + \frac{1}{3} \right ) ^{2} }$

$= \sqrt{ \left [ N^{2} - 2 \cdot N \cdot \frac{1}{3} + \left ( \frac{1}{3}\right )^{2} \right ]+ \left [ N ^{2} + 2 \cdot N \cdot \frac{1}{3} + \left ( \frac{1}{3}\right )^{2} \right ] }$

$= \sqrt{ \left ( N^{2} - \frac{2}{3}N + \frac{1}{9} \right ) + \left ( N^{2} + \frac{2}{3}N + \frac{1}{9} \right ) }$

$= \sqrt{ N^{2} + N^{2} - \frac{2}{3}N + \frac{2}{3}N + \frac{1}{9} + \frac{1}{9} }$

$= \sqrt{ 2N^{2} +\frac{2} {9} }$

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Example Question #171 : Lines

Find the length of a line segment whose endpoints are located at the coordinates (7,-4) and (-5,1) .

Possible Answers:

$\sqrt{13}$

$2\sqrt{42}$

$4\sqrt{10}$

Correct answer:

Explanation:

According to the distance formula, the linear distance between two points (x_a, y_a) and (x_b,y_b) is given by

$d=\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$ ,

which is a generalization of the Pythagorean Theorem c^2=a^2+b^2 for any two points on the coordinate plane.

The two points given in this problem are (7,-4) and (-5,1) . Although the question asks for the length of a line segment given its two endpoints, the distance formula is applicable here because the geometric meaning of "distance between two points" is the length of a line segment. Hence, we can use the distance formula to determine the distance between these two endpoints as shown:

$d=\sqrt{((7)-(-5))^2+((-4)-(1))^2}$

$\Rightarrow d=\sqrt{(12)^2+(-5)^2}$

$\Rightarrow d=\sqrt{144+25}$

$\Rightarrow d=\sqrt{169}$

$\Rightarrow d=13$

Hence, the line segment bounded by the points (7,-4) and (-5,1) is units long.

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

One line has four collinear points in order from left to right A, B, C, D. If AB = 10’, CD was twice as long as AB, and AC = 25’, how long is AD?

Possible Answers:

45'

50'

35'

30'

40'

Correct answer:

45'

Explanation:

AB = 10 ’

BC = AC – AB = 25’ – 10’ = 15’

CD = 2 * AB = 2 * 10’ = 20 ’

AD = AB + BC + CD = 10’ + 15’ + 20’ = 45’

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Example Question #1 : Midpoint Formula

The midpoint of line segment AB is (2, -5). If the coordinates of point A are (4, 4), what are the coordinates of B?

Possible Answers:

(6, 14)

(3, -0.5)

(0, -14)

(6, 13)

(0, -13)

Correct answer:

(0, -14)

Explanation:

The fastest way to find the missing endpoint is to determine the distance from the known endpoint to the midpoint and then performing the same transformation on the midpoint. In this case, the x-coordinate moves from 4 to 2, or down by 2, so the new x-coordinate must be 2-2 = 0. The y-coordinate moves from 4 to -5, or down by 9, so the new y-coordinate must be -5-9 = -14.

An alternate solution would be to substitute (4,4) for (x₁,y₁) and (2,-5) for (x,y) into the midpoint formula:

x=(x₁+x₂)/2

y=(y₁+y₂)/2

Solving each equation for (x₂,y₂) yields the solution (0,-14).

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Example Question #1 : How To Find The Endpoints Of A Line Segment

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

Possible Answers:

(29, –15)

(8.5, –2)

None of the other answers

(12, –11)

(22, –9)

Correct answer:

(29, –15)

Explanation:

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

We need to use our generalized midpoint formula:

MP = ( (5 + x)/2, (7 + y)/2 )

Solve each separately:

(5 + x)/2 = 17 → 5 + x = 34 → x = 29

(7 + y)/2 = –4 → 7 + y = –8 → y = –15

Therefore, B is (29, –15).

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Example Question #3 : Midpoint Formula

Line segment AB has an endpoint, A, located at (10,-1) , and a midpoint at (10,0) . What are the coordinates for point B of segment AB?

Possible Answers:

The second endpoint cannot exist

(10,-0.5)

(5,1)

(10,1)

(5,-0.5)

Correct answer:

(10,1)

Explanation:

With an endpoint A located at (10,-1), and a midpoint at (10,0), we want to add the length from A to the midpoint onto the other side of the segment to find point B. The total length of the segment must be twice the distance from A to the midpoint.

A is located exactly one unit below the midpoint along the y-axis, for a total displacement of (0,1). To find point B, we add (10+0, 0+1), and get the coordinates for B: (10,1).

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Example Question #4 : Midpoint Formula

Solve each problem and decide which is the best of the choices given.

What is the distance between the points (1,6) and (-7,3) on a standard coordinate plane?

Possible Answers:

$\sqrt{55}$

$\sqrt{73}$

$\sqrt{39}$

Correct answer:

$\sqrt{73}$

Explanation:

Make a triangle. The points are 8 units apart on the -axis, and units apart on the -axis. Then use the Pythagorean Theorem to find the distance of the hypotenuse, which ends up being $\sqrt{73}$ .

Another way to solve this problem is to use the distance formula,

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

Plugging in the two points we get,

$d=\sqrt{(1--7)^2+(6-3)^2}$

$d=\sqrt{(8)^2+(3)^2}$

$d=\sqrt{64+9}$

$d=\sqrt{73}$

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Example Question #1 : How To Find The Midpoint Of A Line Segment

A line segment has endpoints (0,4) and (5,6). What are the coordinates of the midpoint?

Possible Answers:

(3,9)

(2.5,5)

(2.5,-5)

(0,6)

(0,4)

Correct answer:

(2.5,5)

Explanation:

A line segment has endpoints (0,4) and (5,6). To find the midpoint, use the midpoint formula:

X: (x₁+x₂)/2 = (0+5)/2 = 2.5

Y: (y₁+y₂)/2 = (4+6)/2 = 5

The coordinates of the midpoint are (2.5,5).

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Example Question #2 : How To Find The Midpoint Of A Line Segment

Find the midpoint between (-3,7) and (5,-9)

Possible Answers:

(4,-1)

(1,-8)

(4,-8)

(-1,-1)

(1,-1)

Correct answer:

(1,-1)

Explanation:

You can find the midpoint of each coordinate by averaging them. In other words, add the two x coordinates together and divide by 2 and add the two y coordinates together and divide by 2.

x-midpoint = (-3 + 5)/2 = 2/2 = 1

y-midpoint = (7 + -9)/2 = -2/2 = -1

(1,-1)

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SAT Math : Lines

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #19 : Distance Formula

Example Question #20 : Distance Formula

Example Question #171 : Lines

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

Example Question #1 : Midpoint Formula

Example Question #1 : How To Find The Endpoints Of A Line Segment

Example Question #3 : Midpoint Formula

Example Question #4 : Midpoint Formula

Example Question #1 : How To Find The Midpoint Of A Line Segment

Example Question #2 : How To Find The Midpoint Of A Line Segment

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