Coordinate Geometry

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PSAT Math › Coordinate Geometry

Questions 1 - 10

What line is perpendicular to and passes through ?

$y = \frac{5}{3}x - 3$

$y = \frac{-3}{5}x + 2$

$y = \frac{1}{3}x - 5$

$y = \frac{2}{5}x + 4$

$y = \frac{3}{2}x - 2$

Explanation

Convert the given equation to slope-intercept form.

$y=-\frac{3}{5}+3$

The slope of this line is $-\frac{3}{5}$ . The slope of the line perpendicular to this one will have a slope equal to the negative reciprocal.

The perpendicular slope is $\frac{5}{3}$ .

Plug the new slope and the given point into the slope-intercept form to find the y-intercept.

$2 = \frac{5}{3}(3) + b$

So the equation of the perpendicular line is $y = \frac{5}{3}x - 3$ .

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

(0,4)

(0,6)

(2.5,-5)

(2.5,5)

(3,9)

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).

A circle with a radius of five is centered at the origin. A point on the circumference of the circle has an x-coordinate of two and a positive y-coordinate. What is the value of the y-coordinate?

$\sqrt{23}$

$\sqrt{21}$

Explanation

Recall that the general form of the equation of a circle centered at the origin is:

_x_2 + _y_2 = _r_2

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

_x_2 + _y_2 = 52

_x_2 + _y_2 = 25

Now, the question asks for the positive y-coordinate when x = 2. To solve this, simply plug in for x:

22 + _y_2 = 25

4 + _y_2 = 25

_y_2 = 21

y = ±√(21)

Since our answer will be positive, it must be √(21).

A circle has its origin at . The point is on the edge of the circle. What is the radius of the circle?

$r=\sqrt{74}$

$r=2\sqrt{18}$

$r=6\sqrt{2}$

$r=4\sqrt{21}$

There is not enough information to answer this question.

Explanation

The radius of the circle is equal to the hypotenuse of a right triangle with sides of lengths 5 and 7.

$r=\sqrt{74}$

This radical cannot be reduced further.

We have a square with length 2 sitting in the first quadrant with one corner touching the origin. If the square is inscribed inside a circle, find the equation of the circle.

Explanation

If the square is inscribed inside the circle, in means the center of the circle is at (1,1). We need to also find the radius of the circle, which happens to be the length from the corner of the square to it's center.

Now use the equation of the circle with the center and .

We get

A circle exists entirely in the first quadrant such that it intersects the -axis at . If the circle intersects the -axis in at least one point, what is the area of the circle?

$36\pi$

$9\pi$

$144\pi$

$6\pi$

$Not\ enough\ information$

Explanation

We are given two very important pieces of information. The first is that the circle exists entirely in the first quadrant, the second is that it intersects both the - and -axis.

The fact that it is entirely in the first quadrant means that it cannot go past the two axes. For a circle to intersect the -axis in more than one point, it would necessarily move into another quadrant. Therefore, we can conclude it intersects in exactly one point.

The intersection of the circle with must also be tangential, since it can only intersect in one point. We can thus conclude that the circle must have both - and - intercepts equal to 6 and have a center of .

This leaves us with a radius of 6 and an area of:

$\pi *radius^{2}=\pi*6^{2} = 36\pi$

The endpoints of a diameter of circle A are located at points and . What is the area of the circle?

$20\pi$

$40\pi$

$80\pi$

$160\pi$

$12\pi$

Explanation

The formula for the area of a circle is given by A =πr2 . The problem gives us the endpoints of the diameter of the circle. Using the distance formula, we can find the length of the diameter. Then, because we know that the radius (r) is half the length of the diameter, we can find the length of r. Finally, we can use the formula A =πr2 to find the area.

The distance formula is Actmath_7_113_q1

The distance between the endpoints of the diameter of the circle is:

$d=\sqrt{(6-(-2))^{2}+(8-4)^{2}}$

$d=\sqrt{8^{2}+4^{2}}$

$d=\sqrt{64+16}$

$d=\sqrt{80}=4\sqrt{5}$

To find the radius, we divide d (the length of the diameter) by two.

$r=\frac{d}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}$

Then we substitute the value of r into the formula for the area of a circle.

$A=\pi r^{2}=\pi ({2\sqrt{5}})^{2}=20\pi$

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?

45'

40'

50'

35'

30'

Explanation

AB = 10 ’

BC = AC – AB = 25’ – 10’ = 15’

CD = 2 * AB = 2 * 10’ = 20 ’

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

Whast line goes through the points and ?

Explanation

Let $P_{1}=(1,3)$ and $P_{2}=(7,5)$

The slope is geven by: $m = (y_{2} - y_{1}) \div (x_{2} - x_{1})$ so

$m=\frac{1}{3}$

Then we use the slope-intercept form of an equation; so

$b=\frac{8}{3}$

And we convert

$y=\frac{1}{3}x+\frac{8}{3}$

to standard form.

A circle exists entirely in the first quadrant such that it intersects the -axis at . If the circle intersects the -axis in at least one point, what is the area of the circle?

$36\pi$

$9\pi$

$144\pi$

$6\pi$

$Not\ enough\ information$

Explanation

We are given two very important pieces of information. The first is that the circle exists entirely in the first quadrant, the second is that it intersects both the - and -axis.

This leaves us with a radius of 6 and an area of:

$\pi *radius^{2}=\pi*6^{2} = 36\pi$

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