Coordinate Geometry - GED Math

Card 0 of 20

Question

Use the distance formula to calculate the distance between the points $\small \small (4,2)$ and $\small (-7,3)$ .

Answer

The distance between 2 points can be determined using the distance formula:

$\small D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\small D=\sqrt{(4-(-7))^2+(2-3)^2}$

$\small D=\sqrt{(11)^2+(-1)^2}$

$\small D=\sqrt{121+1}=\sqrt{122}$

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Question

Provide your answer in its most simplified form.

Find the distance between the following two points:

Answer

To find the distance we need to use the distance formula:

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

Plug in your x and y values to get:

$d = \sqrt{(7 - (-3))^{2} + (0 - 5)^{2}}$

Combine like terms to get:

$d = \sqrt{(10)^{2} + (-5)^{2}}$

Continue with your order of operations:

$d = \sqrt{100 + 25}$

$d = \sqrt{125}$

Don't forget to simplify if possible:

$d = 5\sqrt{5}$

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Question

Provide your answer in its most simplified form.

Find the distance between these two points:

Answer

For this problem we must use the distance formula:

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

Plug in your x and y values:

$d = \sqrt{((-10) - 3)^{2} + ((-4) - 8)^{2}}$

Combine like terms:

$d = \sqrt{(-13)^{2} + (-12)^{2}}$

Continue your order of operations:

$d = \sqrt{169 +144}$

$d = \sqrt{313}$

This cannot be simplified so you are left with the correct answer.

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Question

Provide your answer in its most simplified form.

Find the distance between the two following points:

Answer

We must use the distance formula to solve this problem:

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

Plug in your x and y values:

$d = \sqrt{(2 - 5)^{2} + (2 - 6)^{2}}$

Combine like terms:

$d = \sqrt{(-3)^{2} + (-4)^{2}}$

Continue with your order of operations

$d = \sqrt{9 + 16}$

$d = \sqrt{25}$

Simplify to get:

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Question

What is the distance between the points and ?

Answer

Write the distance formula.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the points into the formula.

$D=\sqrt{(-5-3)^2+(6-2)^2}$

$D=\sqrt{(-8)^2+(4)^2} = \sqrt{64+16} = \sqrt{80}$

Factor the radical using factors of perfect squares.

$\sqrt{80} = \sqrt{16}\cdot \sqrt5$

The answer is: $4\sqrt{5}$

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Question

What is the distance between the points and ?

Answer

Remember that you can consider your two points as:

and

From this, remember that the distance formula is:

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Now, for your data, this is:

$\sqrt{(22-4)^2+(6-5)^2}$

or

$\sqrt{(18)^2+(1)^2}=\sqrt{324+1}=\sqrt{325}$

You can simplify this value a little. Identify the prime factors of and move any number that appears in a pair of factors from the interior to the exterior of the square root symbol:

$\sqrt{5\cdot5\cdot13}=5\sqrt{13}$

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Question

What is the distance between the two points and ?

Answer

Remember that you can consider your two points as:

and

From this, remember that the distance formula is:

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Now, for your data, this will look like the following. Be very careful with the negative signs:

$\sqrt{(-11-(-10))^2+(-20-15)^2}$

or

$\sqrt{(-1)^2+(-35)^2}=\sqrt{1+1225}=\sqrt{1226}$

is only factorable into and ; therefore, your answer is in its final form already.

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Question

Find the distance from point to .

Answer

Write the formula to find the distance between two points.

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the points into the radical.

$D = \sqrt{(-1-3)^2+(8-1)^2}$

$D = \sqrt{(-4)^2+(7)^2} = \sqrt{16+49} = \sqrt{65}$

The answer is: $\sqrt{65}$

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Question

What is the distance between and ?

Answer

Write the distance formula.

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the points into the equation.

$D = \sqrt{(5-(-2))^2+(1-3)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49+4}$

The answer is: $\sqrt{53}$

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Question

A triangle on a coordinate plane has the following vertices: . What is the perimeter of the triangle?

Answer

Since we are asked to find the perimeter of the triangle, we will need to use the distance formula to find the length of each side. Recall the distance formula:

$\text{Distance}=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Start by finding the distance between the points $(5, 5)\text{ and }(10, 2)$ :

$\text{Distance}=\sqrt{(2-5)^2+(10-5)^2}=\sqrt{9+25}=\sqrt{34}$

Next, find the distance between $(10, 2)\text{ and }(8, -2)$ .

$\text{Distance}=\sqrt{(-2-2)^2+(8-10)^2}=\sqrt{16+4}=\sqrt{20}$

Then, find the distance between $(5, 5)\text{ and }(8, -2)$ .

$\text{Distance}=\sqrt{(-2-5)^2+(8-5)^2}=\sqrt{49+9}=\sqrt{58}$

Finally, add up the lengths of each side to find the perimeter of the triangle.

$\text{Perimeter}=\sqrt{54}+\sqrt{20}+\sqrt{58}=17.92$

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Question

Use distance formula to find the distance between the following two points.

Answer

Use distance formula to find the distance between the following two points.

Distance formula is as follows:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Note that it doesn't matter which point is "1" and which point is "2" just so long as we remain consistent.

So, let's plug and chug.

$d=\sqrt{(43-22)^2+(54-76)^2}=\sqrt{21^2+(-22)^2}=\sqrt{441+484}=\sqrt{925}$

$d=\sqrt{925}=\sqrt{25}\sqrt{37}=5\sqrt{37}$

So, our answer is

$5\sqrt{37}$

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Question

What is the distance between the points and ?

Answer

Recall the distance formula:

$\text{Distance}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Plug in the given points to find the distance between them.

$\text{Distance}=\sqrt{(5-(-5))^2+(1-10)^2}=\sqrt{100+81}=\sqrt{181}$

The distance between those points is $\sqrt{181}$ .

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Question

Find the distance between the points and .

Answer

Find the distance between the points and .

To find the distance between two points, we will use distance formula (clever name). Distance formula can be thought of as a modified Pythagorean Theorem. What distance formula does is essentially treats our two points as the ends of a hypotenuse on a right triangle, then uses the two side lengths to find the hypotenuse.

Distance formula:

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Pythagorean Theorem

If the connection isn't clear, don't worry, we can still solve for distance.

$D=\sqrt{(413-9)^2+(65-17)^2}$

$D=\sqrt{(404)^2+(48)^2}$

$D=\sqrt{165520}$

$D=406.84\approx 407$

So our answer is 407

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Question

Find the length of the line connecting the following points.

Answer

Find the length of the line connecting the following points.

To find the length of a line, use distance formula.

$d=\sqrt{(y_1-y_2)^2+(x_1-x_2)^2}$

What we are really doing is making a right triangle and using Pythagorean Theorem to find the hypotenuse.

Let's plug in our points and find the distance!

$d=\sqrt{(123-47)^2+(99-19)^2}$

$d=\sqrt{(76)^2+(80)^2}$

$d=\sqrt{(5776+6400}=\sqrt{12176}$

$\sqrt{12176}\approx 110$

So our answer is 110

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Question

A line has slope $\frac{3}{4}$ and -intercept . Give its -intercept.

Answer

The -intercept will be a point for some value . We use the slope formula

$\frac{y_{2}- y _{1}}{x_{2}- x _{1}} = m$ ,

setting $m = \frac{3}{4}, x_{1} =0, x_{2} = a, y_{1} = -7, y_{2} =0$ ,

and solving for :

$\frac{0- (-7)}{a-0} = \frac{3}{4}$

$\frac{ 7}{a} = \frac{3}{4}$

$\frac{ 7}{a} \cdot a = \frac{3}{4} \cdot a$

$7= \frac{3}{4} \cdot a$

$\frac{4} {3}\cdot 7= \frac{4} {3}\cdot \frac{3}{4} \cdot a$

$a = \frac{28} {3} = 9 \frac{1}{3}$

The -intercept is $\left ( 9 \frac{1}{3}, 0 \right )$ .

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Question

A line has slope $\frac{3}{4}$ and -intercept . Give its -intercept.

Answer

The -intercept will be the point for some value . We use the slope formula

$\frac{y_{2}- y _{1}}{x_{2}- x _{1}} = m$ ,

setting $m = \frac{3}{4}, x_{1} = 5, x_{2} = 0, y_{1} = 0, y_{2} = b$ ,

and solving for :

$\frac{b-0}{0- 5} = \frac{3}{4}$

$\frac{b }{ - 5} = \frac{3}{4}$

$\frac{b }{ - 5} \cdot (-5) = \frac{3}{4} \cdot (-5)$

$b= \frac{-15}{4} = -3 \frac{3}{4}$

The -intercept is $\left (0, -3 \frac{3}{4} \right )$ .

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Question

Find the midpoint of the line segment that connects the following points:

$\small (-5,-3)\ and\ (3,7)$

Answer

Use the midpoint formula:

$\small mid=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$\small mid=(\frac{-5+3}{2},\frac{-3+7}{2})$

$\small mid=(\frac{-2}{2},\frac{4}{2})$

$\small mid=(-1,2)$

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Question

You are given points and . is the midpoint of $\overline{AD}$ , is the midpoint of $\overline{AC}$ , and is the midpoint of $\overline{BD}$ . Give the coordinates of .

Answer

Repeated application of the midpoint formula, $\left (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ , yields the following:

is the point and is the point $\left (1, 9 \right )$ . is the midpoint of $\overline{AD}$ , so has coordinates

$\left (\frac{8+1}{2}, \frac{1+9}{2} \right )$ , or $\left ( 4.5, 5\right )$ .

is the midpoint of $\overline{AC}$ , so has coordinates

$\left (\frac{8+4.5}{2}, \frac{1+5}{2} \right )$ , or .

is the midpoint of $\overline{BD}$ , so has coordinates

$\left (\frac{6.25+1}{2}, \frac{3+9}{2} \right )$ , or .

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Question

You are given and . is the midpoint of $\overline{AC}$ ; is the midpoint of $\overline{AD}$ . What are the coordinates of ?

Answer

Repeated application of the midpoint formula $\left (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ yields the following:

Since is the midpoint of $\overline{AC}$ , substitute the coordinates of for $x_{1}$ and $y_{1}$ , set equal to the coordinates of , and solve as follows:

$7 = \frac{3+x_{2}}{2}$

$14 =3+x_{2}$

$x_{2} = 11$

$-1= \frac{7+y_{2}}{2}$

$-2= 7+y_{2}$

$-9= y_{2}$

is the point .

We can find the coordinates of similarly using those of and :

$11= \frac{3+x_{2}}{2}$

$22 = 3+x_{2}$

$19= x_{2}$

$-9= \frac{7+y_{2}}{2}$

$-18= 7+y_{2}$

$-25= y_{2}$

is the point

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Question

What is the midpoint between and ?

Answer

Write the formula to find the midpoint.

$M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substitute the points into the equation.

$M = (\frac{2+1}{2}, \frac{6+(-9)}{2})$

The midpoint is located at: $(\frac{3}{2}, -\frac{3}{2})$

The answer is: $(\frac{3}{2}, -\frac{3}{2})$

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