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SAT II Math II : Coordinate Geometry

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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #131 : Geometry

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

g(x) =-|4x |- 4

g(x) =-|4x+16|

$g(x) =\left | \frac{1}{4}x+1 \right |$

$g(x) =\left | \frac{1}{4}x \right |- 4$

$g(x) =\left | \frac{1}{4}x-1 \right |$

Correct answer:

$g(x) =\left | \frac{1}{4}x \right |- 4$

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph of is translated by shifting each point (x,y) to the point $\left ( x, \frac{1}{4}y \right )$ , the graph of

$g(x) = \frac{1}{4}f(x)$

is formed. If the graph is then shifted down by four units, the new graph is

$g(x) = \frac{1}{4}f(x)- 4$ .

Since the starting graph was f(x) = |x| , the final graph is

$g(x) = \frac{1}{4}|x|- 4$ , or

$g(x) =\left | \frac{1}{4}x \right |- 4$

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Example Question #1 : Transformations

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

$g(x) =\left | \frac{1}{3} x- \frac{2}{3} \right |$

$g(x) =\left | \frac{1}{3} x+ \frac{2}{3} \right |$

$g(x) =\left | \frac{1}{3} x \right |+2$

g(x) = |3x-6|

g(x) = |3x|-2

Correct answer:

$g(x) =\left | \frac{1}{3} x- \frac{2}{3} \right |$

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph of is translated by shifting each point (x,y) to the point $\left ( x, \frac{1}{3}y \right )$ , the graph of

$g(x) = \frac{1}{3}f(x)$

is formed. If the graph is then shifted right by two units, the new graph is

$g(x) = \frac{1}{3}f(x-2)$

Since the starting graph was f(x) = |x| , the final graph is

$g(x) = \frac{1}{3} |x-2|$ , or

$g(x) =\left | \frac{1}{3} x- \frac{2}{3} \right |$

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Example Question #1 : Distance Formula

What is the distance between the point (1,2) and (8,5)?

Possible Answers:

13.9

$\sqrt58$

Correct answer:

$\sqrt58$

Explanation:

For this question we will use the distance formula to solve.

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

In our case (x_1, y_1)=(1,2)

and (x_2, y_2)= (8,5)

Substituting these values in we get the following

$d=\sqrt((8-1)^2+(5-2)^2)$

$d=\sqrt(7^2+3^2)$

$d=\sqrt(49+9)$

$d=\sqrt(58)$

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Example Question #2 : Distance Formula

The points A=(-2,0), B=(0,3), and C=(0,0) makes a triangle. What is the distance between point A and point B?

Possible Answers:

$\sqrt13$

$\sqrt15$

Correct answer:

$\sqrt13$

Explanation:

For this question we need to use the distance formula for points A and B.

Point A will be our (x_1, y_1) and point B will be (x_2, y_2)

Now we substitute these values into the following:

$d=\sqrt((0--2)^2+(3-0)^2)$

$d=\sqrt(2^2+3^2)$

$d=\sqrt(4+9)$

$d=\sqrt13$

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Example Question #3 : Distance Formula

Find the distance between the two points (2,7) and (4,6).

Possible Answers:

$\sqrt{205}$

$\sqrt{5}$

$\sqrt{17}$

$\sqrt{3}$

Correct answer:

$\sqrt{5}$

Explanation:

The distance between two points is found using the formula $d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

For this problem the values are as follows:

$x_{1}=2$

$x_{2}=4$

$y_{1}=7$

$y_{2}=6$

Input the values into the formula and simplify

$d=\sqrt{(4-2)^2+(7-6)^2}=\sqrt{(2)^2+(1)^2}=\sqrt{4+1}=\sqrt{5}$

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Example Question #1 : Midpoint Formula

Find the point halfway between points A and B.

A=(19,54)

B=(67,34)

Possible Answers:

(86,88)

(42,45)

(43,44)

(44,43)

Correct answer:

(43,44)

Explanation:

Find the point halfway between points A and B.

A=(19,54)

B=(67,34)

We are going to need to use midpoint formula. If you ever have difficulty recalling midpoint formula, try to recall that it is basically taking two averages. One average is the average of your x values, the other average is the average of your y values.

$Mid=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Now we plug and chug!

$Mid=(\frac{19+67}{2},\frac{54+34}{2})=(\frac{86}{2},\frac{88}{2})=(43,44)$

So our answer is (43,44)

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Example Question #2 : Midpoint Formula

What is the coordinates of the point exactly half way between (-2, -3) and (5, 7)?

Possible Answers:

$\left(2, \frac{-3}{2}\right)$

(2, -3)

(3,-2)

$\left(\frac{3}{2}, 2\right)$

Correct answer:

$\left(\frac{3}{2}, 2\right)$

Explanation:

We need to use the midpoint formula to solve this question.

$midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

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

and (x_2, y_2)=(5,7)

Therefore, substituting these values in we get the following:

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

$midpoint= (\frac{3}{2}, \frac{4}{2})$

$(\frac{3}{2}, 2)$

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Example Question #3 : Midpoint Formula

Find the midpoint between (2,3) and (-1,9) .

Possible Answers:

$(\frac{3}{2}, -3)$

$(\frac{1}{2}, -6)$

$(\frac{1}{2}, 6)$

$(\frac{3}{2}, 3)$

$(\frac{1}{2}, -3)$

Correct answer:

$(\frac{1}{2}, 6)$

Explanation:

Write the midpoint formula.

$M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substitute the points.

$M = (\frac{2+(-1)}{2}, \frac{3+9}{2}) = (\frac{1}{2}, 6)$

The answer is: $(\frac{1}{2}, 6)$

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Example Question #1 : Other Coordinate Geometry

On the coordinate plane, two lines intersect at the origin. One line passes through the point (3, 2) ; the other, (3, 4) .

Give the measures of the acute angles they form at their intersection (nearest degree).

Possible Answers:

$34 ^{\circ }$

$71^{\circ }$

$21^{\circ }$

$19 ^{\circ }$

$69^{\circ }$

Correct answer:

$19 ^{\circ }$

Explanation:

If $\theta$ is the measure of the angles that two lines with slopes $m_{1}$ and $m_{2}$ form, then

$\tan \theta = \frac{m_{1}-m _{2}}{1+m_{1}m_{2}}$ ,

The slopes of the lines can be found by applying the slope formula

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

using the known points.

For the first line, set $x_{1}= y_{1} = 0, x_{2}= 3, y_{2} = 4$ :

$m_{1} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{4-0}{3-0} = \frac{4}{3}$

The inverse tangent of this is

$\tan^{-1} \frac{4}{3} \approx 53.1^{\circ }$ ,

making this the angle this line forms with the -axis.

For the second line, set $x_{1}= y_{1} = 0, x_{2}= 3, y_{2} = 2$ :

$m_{2} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{2-0}{3-0} = \frac{2}{3}$

The inverse tangent of this is

$\tan^{-1} \frac{2}{3} \approx 33.7^{\circ }$

making this the angle this line forms with the -axis.

Subtract:

Taking the inverse tangent:

$\theta \approx 53.1^{\circ } - 33.7^{\circ } \approx 19.4 ^{\circ }$ .

Rounding to the nearest degree, this is $19 ^{\circ }$ .

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Example Question #2 : Other Coordinate Geometry

In the figure below, regular hexagon ABCDEF has a side length of . Find the y-coordinate of point .

Possible Answers:

3.46

3.99

3.38

4.09

Correct answer:

3.46

Explanation:

From the given information, we know that the coordinate for must be (0, 8) .

Recall that the interior angle of a regular hexagon is $120^{\circ}$ . Thus, we can draw in the following 30-60-90 triangle.

Since we know that this is a 30-60-90 triangle, we know that the sides must be as marked, in the ratio of $1:\sqrt3:2$ . Thus, the y-coordinate of must be 3.46 .

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All SAT II Math II Resources

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SAT II Math II : Coordinate Geometry

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

All SAT II Math II Resources

Example Questions

Example Question #131 : Geometry

Example Question #1 : Transformations

Example Question #1 : Distance Formula

Example Question #2 : Distance Formula

Example Question #3 : Distance Formula

Example Question #1 : Midpoint Formula

Example Question #2 : Midpoint Formula

Example Question #3 : Midpoint Formula

Example Question #1 : Other Coordinate Geometry

Example Question #2 : Other Coordinate Geometry

All SAT II Math II Resources

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