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Intermediate Geometry : Coordinate Geometry

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Example Questions

Example Question #1 : How To Find The Equation Of A Curve

Which equation has a y-intercept at 2 and x-intercepts at -1 and 6?

Possible Answers:

$\small y = -3(x+1)(x-2)$

$\small y = \frac{-1}{3}(x+1)(x-6)$

$\small y = \frac{1}{3}(x-1)(x+6)$

$\small y=3 (x+1)(x-2 )$

$\small \small y = (x+1)(x-6)$

Correct answer:

$\small y = \frac{-1}{3}(x+1)(x-6)$

Explanation:

In order for the equation to have x-intercepts at -1 and 6, it must have $\small (x+1)$ and (x-6) as factors. This leaves us with only 2 choices, $\small \small y = (x+1)(x-6)$ or $\small y = \frac{-1}{3}(x+1)(x-6)$

This equation must also have a y-intercept of 2. This means that plugging in 0 for x will gives us a y-value of 2. Because we have two options, we could plug in 0 for x in each to see which gives us an answer of 2:

a) $\small y = (0+1)(0-6) = 1*-6 = -6$ we can eliminate that choice

b) $\small y= -\frac{1}{3}(0+1)(0-6) = -\frac{1}{3}*1*-6 = 2$ this must be the right choice.

If we hadn't been given multiple options, we could have set up the following equation to figure out the third factor:

$\small 2 = C(0+1)(0-6)$

$\small 2 = C*1*-6$

$\small 2 = -6C$ divide by -6

$\small -\frac{1}{3} = C$

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Example Question #2 : How To Find The Equation Of A Curve

Which equation would have an x-intercept at $\small x=2$ and a y-intercept at $\small y=-3$ ?

Possible Answers:

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

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

$\small y = \frac{3}{2}x + 3$

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

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

Correct answer:

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

Explanation:

We're writing the equation for a line passing through the points $\small (0,-3)$ and $\small (2,0)$ . Since we already know the y-intercept, we can figure out the slope of this line and then write a slope-intercept equation.

To determine the slope, divide the change in y by the change in x:

$\small \frac{-3 - 0 }{ 0 - 2 } = \frac{-3}{-2} = \frac{3}2$

The equation for this line would be $\small y = \frac{3}{2}x - 3$ .

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Example Question #2 : How To Find The Equation Of A Curve

Write the equation of a line with intercepts (0, 4) and (-1, 0 )

Possible Answers:

y = 4x + 4

$y = -\frac{1}{4}x + 1$

y = 5x-1

y = -x+5

$y = \frac{1}{4}x +1$

Correct answer:

y = 4x + 4

Explanation:

The line will eventually be in the form y = mx+b where is the y-intercept.

The y-intercept in this case is .

To find the equation, plug in for , and the other point, (-1,0) as x and y:

0 = m(-1)+ 4

0 = -m + 4 add to both sides

m = 4

This means the equation is y = 4x + 4

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Example Question #2 : How To Find The Equation Of A Curve

Which equation has the x- and y-intercepts (2,0) and (0,7) ?

Possible Answers:

$y = \frac{2}{7} x + 2$

$y = \frac{7}{2} x +3$

$y = -\frac{2}{7} + 3$

$y = -\frac{7}{2}x + 7$

y = 5x + 1

Correct answer:

$y = -\frac{7}{2}x + 7$

Explanation:

Since this line has the y-intercept (0,7)

, we know that in the form y = mx+b

We can plug in the other intercept's coordinates for

and

to solve for

subtract

divide by

$\frac{-7}{2} = m$

The line is $y = -\frac{7}{2}x + 7$

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Example Question #1 : How To Find The Equation Of A Circle

What is the equation for a circle centered at (1,2) with a radius of ?

Possible Answers:

$(x+2)^{2}+ (y+1)^{2} = 36$

$(x+1)^{2}+ (y-2)^{2} = 9$

$(x-2)^{2}+ (y+1)^{2} = 18$

$(x-1)^{2}+ (y+2)^{2} = -9$

$(x-1)^{2}+ (y-2)^{2} = 9$

Correct answer:

$(x-1)^{2}+ (y-2)^{2} = 9$

Explanation:

The general equation for a circle centered at the origin is given by $x^{2}+y^{2}=r^{2}$ where is the radius of the circle.

To translate the origin to the first quadrant we need to subtract the appropriate amount to bring it back to center.

So the equation becomes $(x-1)^{2}+(y-2)^{2}=9$

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Example Question #1 : How To Find The Equation Of A Circle

A circle has a diameter starting at (3,2) and ending at (9,10) . What is the equation of the circle?

Possible Answers:

$(x-4)^{2}+(y-8)^{2}=25$

$(x-6)^{2}+(y-6)^{2}=25$

$(x+6)^{2}-(y+6)^{2}=25$

$(x-6)^{2}+(y-6)^{2}=100$

$(x-6)^{2}+(y-6)^{2}=5$

Correct answer:

$(x-6)^{2}+(y-6)^{2}=25$

Explanation:

Using the distance formula, the diameter is 10 units long, so the radius is 5. Then, by finding the midpoint of the diameter, you know the center of the cirle. Plugging values into the circle equation yields the final answer.

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Example Question #3 : Circles

What is the equation of a circle with diameter , that is tangent to the x-axis at the point (3,0) and the y-axis at the point (0,3) ?

Possible Answers:

(x+3)^2+(y+3)^2=3

(x-3)^2+(y-3)^2=3

(x+3)^2+(y+3)^2=9

(x-3)^2+(y-3)^2=9

(x+3)^2-(y+3)^2=9

Correct answer:

(x-3)^2+(y-3)^2=9

Explanation:

The equation for a circle is (x-h)^2+(y-k)^2=r^2 where the center of the circle lies at the point and the radius of the circle is . If we are looking for a circle with a diameter of , then its radius must be . For the circle to be tangent to the x-axis at the point (3,0) and the y-axis at (0,3) , it must be centered at the point (3,3) . Therefore, the equation of the circle will be (x-3)^2+(y-3)^2=9 .

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Example Question #3 : How To Find The Equation Of A Circle

A circle has its center at the point (3,7) and a radius of units.

What is the equation of the circle?

Possible Answers:

(x-3)^2+(y-7)^2=4

x-3+y-7=16

(x+3)^2+(y+7)^2=16

(x-3)^2+(y-7)^2=16

Correct answer:

(x-3)^2+(y-7)^2=16

Explanation:

The equation for a circle is

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

where (h, k) is the center of the circle and r is the radius.

Plugging in the values of h=3 , k=7 , and r=4 , we get

(x-3)^2+(y-7)^2=16

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Example Question #2 : Circles

Find the equation of a circle if the radius of the circle is and the center is located at the origin.

Possible Answers:

$x^2+y^2 = \sqrt2$

$x^2+y^2 = \frac{1}{2}$

2x^2+y^2 = 1

x^2+y^2 = 2

x^2+y^2 = 4

Correct answer:

x^2+y^2 = 4

Explanation:

The formula for the equation of a circle is:

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

The values of $\left ( h,k\right )$ represent the center, and both values are zero at the origin.

Plug in the known values and reduce.

(x-0)^2+(y-0)^2 = 2^2

x^2+y^2 = 4

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Example Question #1 : Circles

Write the equation for a circle with the domain of [2,8] and a range of [-1,5] .

Possible Answers:

$(x-2)^{2}+(y-8)^{2}=9$

$(x+1)^{2}+(y-8)^{2}=36$

$(x-5)^{2}+(y-2)^{2}=9$

$(x-6)^{2}+(y-6)^{2}=36$

Correct answer:

$(x-5)^{2}+(y-2)^{2}=9$

Explanation:

The domain of a circle gives the width of the circle, the range of a circle gives the height of a circle - both of which equal the diameter.

The domain and range numbers are both 6 apart, meaning that the diameter of the circle is 6, which means the radius of the circle is 3.

The halfway point between the domain gives the x-coordinate for the center of the circle, which in this case is 5. 5 is halfway between 2 and 8.

The halfway point between the range gives the y-coordinate for the center of the circle, which in this case is 2. 2 is halfway between -1 and 5.

Now we have all of the information needed to plug into the equation for a circle in standard form as shown below where r is the radius and the center is (h,k):

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

Given a circle with radius of 3, and center of (5,2) we get the below equation of a circle.

$(x-5)^{2}+(y-2)^{2}=9$

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Intermediate Geometry : Coordinate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #1 : How To Find The Equation Of A Curve

Example Question #2 : How To Find The Equation Of A Curve

Example Question #2 : How To Find The Equation Of A Curve

Example Question #2 : How To Find The Equation Of A Curve

Example Question #1 : How To Find The Equation Of A Circle

Example Question #1 : How To Find The Equation Of A Circle

Example Question #3 : Circles

Example Question #3 : How To Find The Equation Of A Circle

Example Question #2 : Circles

Example Question #1 : Circles

All Intermediate Geometry Resources

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