Coordinate Geometry

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SAT Math › Coordinate Geometry

Questions 1 - 10

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:

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

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.

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.

What is the slope of a line which passes through coordinates $\dpi{100} \small (3,7)$ and $\dpi{100} \small (4,12)$ ?

$\dpi{100} \small 5$

$\dpi{100} \small \frac{1}{5}$

$\dpi{100} \small \frac{1}{2}$

$\dpi{100} \small 2$

$\dpi{100} \small 3$

Explanation

Slope is found by dividing the difference in the $\dpi{100} \small y$ -coordinates by the difference in the $\dpi{100} \small x$ -coordinates.

$\dpi{100} \small \frac{(12-7)}{(4-3)}=\frac{5}{1}=5$

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).

There is a line defined by the equation below:

There is a second line that passes through the point and is parallel to the line given above. What is the equation of this second line?

$y=-\frac{3}{4}x+2.75$

$y=-\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.75$

$y=\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.625$

Explanation

Parallel lines have the same slope. Solve for the slope in the first line by converting the equation to slope-intercept form.

3x + 4y = 12

4y = _–_3x + 12

y = –(3/4)x + 3

slope = _–_3/4

We know that the second line will also have a slope of _–_3/4, and we are given the point (1,2). We can set up an equation in slope-intercept form and use these values to solve for the y-intercept.

y = mx + b

2 = _–_3/4(1) + b

2 = _–_3/4 + b

b = 2 + 3/4 = 2.75

Plug the y-intercept back into the equation to get our final answer.

y = –(3/4)x + 2.75

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

Consider the lines described by the following two equations:

4y = 3x2

3y = 4x2

Find the vertical distance between the two lines at the points where x = 6.

Explanation

Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.

There is a line defined by the equation below:

There is a second line that passes through the point and is parallel to the line given above. What is the equation of this second line?

$y=-\frac{3}{4}x+2.75$

$y=-\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.75$

$y=\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.625$

Explanation

Parallel lines have the same slope. Solve for the slope in the first line by converting the equation to slope-intercept form.

3x + 4y = 12

4y = _–_3x + 12

y = –(3/4)x + 3

slope = _–_3/4

We know that the second line will also have a slope of _–_3/4, and we are given the point (1,2). We can set up an equation in slope-intercept form and use these values to solve for the y-intercept.

y = mx + b

2 = _–_3/4(1) + b

2 = _–_3/4 + b

b = 2 + 3/4 = 2.75

Plug the y-intercept back into the equation to get our final answer.

y = –(3/4)x + 2.75

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