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SAT II Math I : Coordinate Geometry

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #462 : Advanced Geometry

What is the period of the function?

Screen_shot_2013-07-16_at_10.04.45_am

Possible Answers:

3π

4π

2π

Correct answer:

4π

Explanation:

The period is the time it takes for the graph to complete one cycle.

In this particular case we have a sine curve that starts at 0 and completes one cycle when it reaches $4\pi$ .

Therefore, the period is $4\pi$

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Example Question #21 : How To Find Transformation For An Analytic Geometry Equation

Assume we have a triangle, $\bigtriangleup ABC$ , with the following vertices:

(4,6) , (2,9) , and (5,10)

If $\bigtriangleup ABC$ were reflected across the line y=x , what would be the coordinates of the new vertices?

Possible Answers:

$(6,4),\ (9,2),\ \textup{and } (10,5)$

$(6, -4),\ (9, -2),\ \textup{and } (10, -5)$

$(4, -6),\ (2, -9),\ \textup{and }(5, -10)$

$(-4, 6),\ (-2, 9),\ \textup{and } (-5, 10)$

$(-6,-4),\ (-9,-2),\ \textup{and } (-10, -5)$

Correct answer:

$(6,4),\ (9,2),\ \textup{and } (10,5)$

Explanation:

When we reflect a point across the line, y=x , we swap the x- and y-coordinates; therefore, in each point, we will switch the x and y-coordinates:

(4,6) becomes (6,4) ,

(2,9) becomes (9,2) , and

(5, 10) becomes (10,5) .

The correct answer is the following:

$(6,4),\ (9,2),\ \textup{and } (10,5)$

The other answer choices are incorrect because we only use the negatives of the coordinate points if we are flipping across either the x- or y-axis.

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Example Question #1 : Distance Formula

Find the distance between (3, 4) and (-12, -5) .

Possible Answers:

$3 \sqrt {34}$

$2 \sqrt {6}$

$\sqrt {82}$

$10 \sqrt {3}$

Correct answer:

$3 \sqrt {34}$

Explanation:

For this problem we will need to use the distance formula:

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

In our case,

(x_1,y_1)=(-12, -5) and (x_2,y_2)=(3, 4) .

Plugging these values into the formula we are able to find the distance.

$d = \sqrt {(3-(-12))^2 + (4-(-5)))^2))}$

$=\sqrt {15^2+9^2}$

$= \sqrt {306}$

$= \sqrt {9 \cdot 34} = 3 \sqrt {34}$

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Example Question #2 : Distance Formula

What is the distance between (1,6) and (-3,-2) ?

Possible Answers:

$4\sqrt2$

$\frac{4}{3}$

$2\sqrt5$

$5\sqrt4$

$4\sqrt5$

Correct answer:

$4\sqrt5$

Explanation:

Write the distance formula.

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

Substitute the points:

(x_1,y_1)=(1,6)

(x_2,y_2)=(-3,-2)

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

$d=\sqrt{16+64} = \sqrt{80}$

The radical can be broken down into factors of perfect squares.

$\sqrt{80} = \sqrt{16}\cdot \sqrt5 = 4\sqrt5$

The answer is: $4\sqrt5$

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Example Question #1 : Midpoint Formula

Find the midpoint of the line that passes through the points (-1,4) and (5,2) .

Possible Answers:

(2,3)

(2,-3)

(3,3)

(3,-2)

(-3,2)

Correct answer:

(2,3)

Explanation:

Recall the midpoint formula as $(\frac{x1+x2}{2}, \frac{y1+y2}{2})$ .

Thus,

$(\frac{-1+5}{2},\frac{4+2}{2}) = (2,3)$

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Example Question #602 : Sat Subject Test In Math I

Find the midpoint of the line that passes through the points (2,-1) and (-2,1) .

Possible Answers:

(2,1)

(2,-1)

(-2,-1)

(-2,1)

(0,0)

Correct answer:

(0,0)

Explanation:

Recall the midpoint formula as $(\frac{x1+x2}{2}, \frac{y1+y2}{2})$ .

Thus,

$(\frac{2+(-2)}{2},\frac{-1+1}{2}) = (0,0)$

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Example Question #603 : Sat Subject Test In Math I

What is the midpoint between $\left ( 2,3\right )$ and $\left ( -7,4\right )$ ?

Possible Answers:

(-5,-7)

$\left(\frac{-5}{2}, \frac{7}{2}\right)$

(7,5)

(-5,7)

Correct answer:

$\left(\frac{-5}{2}, \frac{7}{2}\right)$

Explanation:

The formula to find the midpoint is as follows:

$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

In our case our (x_1, y_1)= (2, 3)

our (x_2, y_2)=(-7, 4)

substituting in these values we get

midpoint = $\left(\frac{2+-7}{2}, \frac{3+4}{2}\right)=\left(\frac{-5}{2}, \frac{7}{2}\right)$

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Example Question #4 : Midpoint Formula

What is the midpoint between $\left ( 1,3\right )$ and $\left ( 5,9\right )$ ?

Possible Answers:

$\left ( 1,5\right )$

$\left ( 1,9\right )$

$\left ( 3,9\right )$

$\left ( 3,5\right )$

$\left ( 3,6\right )$

Correct answer:

$\left ( 3,6\right )$

Explanation:

To find the midpoint of two points you find the average of both the values and values.

For ,

$\frac{1+5}{2}=3$ .

For ,

$\frac{3+9}{2}=6$ .

This means the midpoint is $\left ( 3,6\right )$ .

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

What is the midpoint of the points (3,12) and (9,15)?

Possible Answers:

(8,12)

(7,13)

(1.5,7.5)

(6,13.5)

Correct answer:

(6,13.5)

Explanation:

To find the midpoint we must know the midpoint formula which is

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

We then take the -coordinate from the first point and plug it into the formula as $x_{1}$ .

We take the -coordinate from the second point and plug it into the formula as $x_{2}$ .

We then do the same for $y_{1}$ and $y_{2}$ .

With all of the points plugged in our equation will look like this.

$(\frac{3+9}{2},\frac{12+15}{2})$

We then perform the necessary addition and division to get the answer of

$(\frac{12}{2},\frac{27}{2})=(6,13.5)$

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

Find the midpoint of the line segment that connects the two points below.

Point 1: (0,-3)

Point 2: (-2, 4)

Possible Answers:

$(-1, \frac{7}{2})$

$(1, -\frac{1}{2})$

$(-1,-\frac{1}{2})$

$(1, -\frac{7}{2})$

$(-1, \frac{1}{2})$

Correct answer:

$(-1, \frac{1}{2})$

Explanation:

The average of the the -coordinates and the average of the y-coordinates of the given points will give you the mid-point of the line that connects the points.

$\textup{midpoint}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ , where (x_1,y_1) is (0,-3) and (x_2,y_2) is (-2, 4) .

$(\frac{(0)+(-2)}{2}, \frac{(-3)+(4)}{2})$

$(\frac{-2}{2},\frac{1}{2})$

$(-1, \frac{1}{2})$

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SAT II Math I : Coordinate Geometry

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #462 : Advanced Geometry

Example Question #21 : How To Find Transformation For An Analytic Geometry Equation

Example Question #1 : Distance Formula

Example Question #2 : Distance Formula

Example Question #1 : Midpoint Formula

Example Question #602 : Sat Subject Test In Math I

Example Question #603 : Sat Subject Test In Math I

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

All SAT II Math I Resources

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