SAT Math - Coordinate Geometry

Give the area of the triangle on the coordinate plane that is bounded by the axes and the line of the equation .

$56 \frac{1}{4}$

$112\frac{1}{2}$

Explanation

It is necessary to find the vertices of the triangle, each of which is a point at which two of the three lines intersect.

The two axes intersect at the origin, making this one vertex.

The other two points of intersection are the intercepts of the line of equation . Since this equation is in slope-intercept form , where is the -coordinate of the -intercept, then , and the -intercept is the point . The -coordinate of the -intercept, , can be found by setting and solving for :

$a = \frac{15}{2} = 7 \frac{1}{2}$ .

The three vertices are located at $(0,0), \left ( 7\frac{1}{2}, 0 \right ), (0. -15)$

The line in question is shown below, with the bounded triangle shaded in:

Triangle z

The lengths of its legs are equal to the absolute values of the nonzero coordinates of its two intercepts - $a' = 7 \frac{1}{2}, b' = 15$ . The area of this right triangle is half their product:

$A = \frac{1}{2} a 'b' = \frac{1}{2} \cdot 7\frac{1}{2} \cdot 15 = 56 \frac{1}{4}$ .

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.

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 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 square on the coordinate plane has vertices at the points with coordinates , , , and . Give the equation of the circle that circumscribes the square.

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{4}$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} =81$

$\left ( x -9 \right )^{2} + \left ( y- 9\right )^{2} = \frac{81}{2}$

$\left ( x -9 \right )^{2} + \left ( y- 9\right )^{2} = 81$

Explanation

The equation of the circle on the coordinate plane with radius and center is

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

The figure referenced is below:

Circle x

The center of the circle is at the point of intersection of the diagonals, which, as is the case with any rectangle, bisect each other. Therefore, looking at the diagonal with endpoints and , we can set $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the midpoint formula:

$h =\frac{ x_{1}+ x_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

and

$k=\frac{ y_{1}+ y_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

The center of the circumscribing circle is therefore $\left ( \frac{9}{2}, \frac{9}{2} \right )$ .

The radius of the circumscribing circle is the distance from this point to any point on the circle. The distance formula can be used here:

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

Since we are actually trying to find $r ^{2}$ , we will use the form

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

Choosing the radius with endpoints and $\left ( \frac{9}{2}, \frac{9}{2} \right )$ , we set $x_{1} = y_{1} = 0 , x_{2} = y_{2} = \frac{9}{2}$ and substitute:

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

$= \left ( \frac{9}{2} \right )^{2} + \left ( \frac{9}{2} \right )^{2}$

$= \frac{81}{4 } + \frac{81}{4 }$

$= \frac{81}{2}$

Setting $h =k= \frac{9}{2}$ and $r^{2}= \frac{81}{2}$ and substituting in the circle equation:

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

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$ , the correct response.

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.

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

A square on the coordinate plane has as its vertices the points with coordinates , , , and . Give the equation of the circle inscribed inside this square.

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{4 }$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{2 }$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = 81$

$\left ( x- 9 \right ) ^{2} + \left ( y-9 \right )^{2} = \frac{81}{4 }$

$\left ( x- 9 \right ) ^{2} + \left ( y-9 \right )^{2} = 81$

Explanation

The equation of the circle on the coordinate plane with radius and center is

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

The figure referenced is below:

Incircle 1

The center of the inscribed circle is the center of the square, which is where its diagonals intersect; this point is the common midpoint of the diagonals. The coordinates of the midpoint of the diagonal with endpoints at and can be found by setting $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the following midpoint formulas:

$x_{M} = \frac{x_{1}+ x_{2}}{2} = \frac{0+9}{2} = \frac{9}{2}$

$y_{M} = \frac{y_{1}+ y_{2}}{2} = \frac{0+9}{2} = \frac{9}{2}$

This point, $\left (\frac{9}{2}, \frac{9}{2} \right )$ , is the center of the circle. The radius can easily be seen to be half the length of one side; each side is 9 units long, so the radius is half this, or $\frac{9}{2}$ .

Setting $h = k = r = \frac{9}{2}$ in the circle equation:

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

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \left (\frac{9}{2} \right ) ^{2}$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{4 }$

Find the midpoint of the line that passes through the points and .

Explanation

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

Thus,

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

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

On the coordinate plane, , , and are the points with coordinates , , and , respectively. Lines , , and are the perpendicular bisectors of $\overline{XY}$ , $\overline{XZ}$ , and $\overline{YZ}$ , respectively.

and intersect at a point ; and intersect at a point ; and intersect at a point .

Which of these statements is true of , , and ?

, , and are the same point.

, , and are distinct and collinear.

, , and are distinct and are the vertices of a triangle similar to $\bigtriangleup XYZ$ .

and are the same point; is a different point.

, , and are distinct and are the vertices of an equilateral triangle.

Explanation

Another way of viewing this problem is to note that the three given vertices form a triangle $\bigtriangleup XYZ$ whose sides' perpendicular bisectors intersect at the points , , and . However, the three perpendicular bisectors of the sides of any triangle always intersect at a common point. The correct response is that , , and are the same point.

Coordinate Geometry

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