Coordinate Geometry - SAT Subject Test in Math II

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Question

Refer to the above figure. The circle has its center at the origin. What is the equation of the circle?

Answer

The equation of a circle with center and radius is

$(x-h)^{2}+ \left ( y - k\right )^{2} = r^{2}$

The center is at the origin, or , so . To find $r^{2}$ , use the distance formula as follows:

$r^{2} = \left ( 6-0\right )^{2} +\left ( 8 -0 \right )^{2} = 6^{2}+8^{2} = 36 + 64 = 100$

Note that we do not actually need to find .

We can now write the equation of the circle:

$(x-0)^{2}+ \left ( y - 0\right )^{2} = 100$

$x^{2}+y^{2} = 100$

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Question

Refer to the above diagram. The circle has its center at the origin; is the point . What is the length of the arc $\widehat{AB}$ , to the nearest tenth?

Answer

First, it is necessary to determine the radius of the circle. This is the distance between and , so we apply the distance formula:

$r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}$

$r = \sqrt{\left ( -9-0\right )^{2} +\left ( -12 -0 \right )^{2} }$

$r=\sqrt{ (-9)^{2}+(-12)^{2}}$

$r= \sqrt{81+144 }= \sqrt{225} = 15$

The circumference of the circle is

$C = 2 \pi r = 2 \pi \cdot 15 = 30 \pi$

Now we need to find the degree measure of the arc. We can do this best by examining this diagram:

The degree measure of $\widehat{AB}$ is also the measure $\theta$ of the central angle formed by the green radii. This is found using the relationship

$\tan \theta = \frac{y}{x}$

$\tan \theta = \frac{-12}{-9} \approx 1.333$

Using a calculator, we find that $\tan ^{-1} 1.3333 \approx 53.12^{\circ }$ . We can adjust for the location of :

$\theta \approx \left (180 - 53.12 \right )^{\circ } \approx 126.87^{\circ }$

which is the degree measure of the arc.

Now we can calculate the length of the arc:

$L \approx \frac{126.87}{360} \cdot C$

$L \approx \frac{126.87}{360} \cdot 30 \pi \approx 33.2$

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Question

A triangle has its vertices at the points with coordinates , , and . Give the equation of the circle that circumscribes it.

Answer

The circumscribed circle of a triangle is the circle which passes through all three vertices of the triangle.

In general form, the equation of a circle is

$x^{2}+ y^{2}+ Ax + By + C = 0$ .

Since the circle passes through the origin, substitute ; the equation becomes

$0^{2}+0^{2}+ A \cdot 0 + B \cdot 0 + C = 0$

Therefore, we know the equation of any circle passing through the origin takes the form

$x^{2}+ y^{2}+ Ax + By = 0$

for some .

Since the circle passes through , substitute ; the equation becomes

$7^{2}+ 0^{2}+ A \cdot 7 + B \cdot 0 = 0$

Solving for :

Now we know that the equation takes the form

$x^{2}+ y^{2}-7x + By = 0$

for some .

Since the circle passes through , substitute ; the equation becomes

$4^{2}+ 4^{2}- 7 \cdot 4 + B \cdot 4 = 0$

Solving for :

The general form of the equation of the circle is therefore

$x^{2}+ y^{2}-7x -y = 0$

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Question

On the coordinate plane, the vertices of a square are at the points with coordinates . Give the equation of the circle.

Answer

The figure in question is below.

The center of the circle can be seen to be the origin, so, if the radius is , the equation will be

$x^{2}+ y^{2} = r^{2}$ .

The circle passes through the midpoints of the sides, so we will find one of these midpoints. The midpoint $(x_{m}, y_{m})$ of the segment with endpoints and can be found by using the midpoint equations, setting $x_{1} = y_{2}= 7, x_{2} = y_{1}= 0$ :

$x_{m} = \frac{x_{1}+x_{2}}{2} = \frac{7+0}{2} = \frac{7}{2}$

$y_{m} = \frac{y_{1}+y_{2}}{2} = \frac{0+7 }{2} = \frac{7}{2}$

The circle passed though this midpoint $\left ( \frac{7}{2}, \frac{7}{2} \right )$ . The segment from this point to the origin is a radius, and its length is equal to . Using the following form of the distance formula, since we only need the square of the radius:

$r ^{2} = (x_{2}-x_{1}) ^{2} + (y_{2}-y_{1}) ^{2}$ ,

set $x_{1} = y_{1} = 0, x_{2}= y_{2}= \frac{7}{2}$ :

$r ^{2} =\left (\frac{7}{2}-0\right ) ^{2}+\left (\frac{7}{2}-0\right ) ^{2}$

$=\left (\frac{7}{2} \right ) ^{2}+\left (\frac{7}{2}\right ) ^{2}$

$= \frac{49}{4} + \frac{49}{4}$

$= \frac{49}{2}$

Substituting in the circle equation for $r ^{2}$ , we get the correct response,

$x^{2}+ y^{2}= \frac{49}{2}$

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Question

Find the diameter of the circle with the equation .

Answer

Start by putting the equation in the standard form of the equation of a circle by completing the square. Recall the standard form of the equation of a circle:

, where the center of the circle is at and the radius is .

From the equation, we know that .

$r=\sqrt{16}=4$

Since the radius is , double its length to find the length of the diameter. The length of the diameter is .

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Question

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

Answer

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

and

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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Question

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?

Answer

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

Point A will be our and point B will be

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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Question

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

Answer

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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Question

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

Answer

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

and

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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Question

Find the point halfway between points A and B.

Answer

Find the point halfway between points A and B.

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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Question

Find the midpoint between and .

Answer

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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Question

On the coordinate plane, two lines intersect at the origin. One line passes through the point ; the other, .

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

Answer

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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Question

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

Answer

From the given information, we know that the coordinate for must be .

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

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

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Question

Which of the following equations represent a parabola?

Answer

The parabola is represented in the form . If there is a variable in the denominator or as an exponent, it is not a parabola.

The only equation that has an order of two is: $y=\frac{1}{2}x^2$

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Question

Which of the following symmetries applies to the graph of the relation

$\left (x- 4 \right )^{2} + y ^{2}= 18$ ?

I) Symmetry with respect to the origin

II) Symmetry with respect to the -axis

III) Symmetry with respect to the -axis

Answer

The relation

$(x-h)^{2} + (y - k)^{2} = r^{2}$

is a circle with center and radius .

In other words, it is a circle with center at the origin, translated right units and up units.

$\left (x- 4 \right )^{2} + y ^{2}= 18$

or

$\left (x- 4 \right )^{2} +\left ( y - 0 \right ) ^{2}= 18$

is a circle translated right 4 units and up zero units. The upshot is that the circle moves along the -axis only, and therefore is symmetric with respect to the -axis, but not the -axis. Also, as a consequence, it is not symmetric with respect to the origin.

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Question

Refer to the above figure.

Which of the following functions is graphed?

Answer

Below is the graph of :

The given graph is the graph of shifted 6 units left (that is, unit right) and 3 units up.

The function graphed is therefore

where . That is,

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Question

Refer to the above figure.

Which of the following functions is graphed?

Answer

Below is the graph of :

If the graph of is translated by shifting each point to the point , the graph of

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

Since the starting graph was , the final graph is

, or

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Question

Refer to the above figure.

Which of the following functions is graphed?

Answer

Below is the graph of :

If the graph of is translated by shifting each point 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 , 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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Question

Refer to the above figure.

Which of the following functions is graphed?

Answer

Below is the graph of :

If the graph of is translated by shifting each point 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 , 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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Question

Refer to the above figure.

Which of the following functions is graphed?

Answer

Below is the graph of :

If the graph of is translated by shifting each point to the point , the graph of

is formed. If the graph is then shifted upward by three units, the new graph is

Since the starting graph was , the final graph is

, or,

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