Equations of Lines - Math

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Question

The points A, B, and C reside on a line segment. B is the midpoint of AC. If line AB measures 6 units in length, what is the length of line AC?

Answer

If B is the midpoint of AC, then AC is twice as long as AB. We are told that AB=6.

The diagram shows six units between points A and B, with B as the midpoint of segment AC. Therefore segment BC is also six units long, so line AC is twelve units long.

$\small (2)(6)=12$

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Question

What is the length of a line with endpoints and ?

Answer

The formula for the length of a line is very similiar to the pythagorean theorem:

Plug in our given numbers to solve:

$\sqrt{169}=\sqrt{l^2}$

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Question

What is the length of a line segment with end points and ?

Answer

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

Plug in the given values and solve for the length.

$\sqrt{61}=\sqrt{l^2}$

$\sqrt{61}=l$

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Question

What would be the length of a line with endpoints at and ?

Answer

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

Plug in the given values and solve for the length.

$\sqrt{73}=\sqrt{l^2}$

$\sqrt{73}=l$

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Question

If a line has a length of , and the endpoints are and , what is the value of ?

Answer

The formula for the length of a line, l, is the distance formula, which is very similar to the Pythagorean Theorem.

Note that the problem has already given us a value for the length of the line. That means . Plug in all of the given values and solve for the missing term.

Subtract $\small 16$ from both sides.

$\sqrt{y^2}=\sqrt{9}$

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Question

What line goes through the points and ?

Answer

Find the slope between the two points:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{15-5}{5-(-1)}=\frac{10}{6}=\frac{5}{3}$

Next, use the slope-intercept form of the equation:

or $15=\frac{5}{3}(5)+b$ where $b=\frac{20}{3}$

So the equation becomes $y=\frac{5}{3}x+ \frac{20}{3}$ or in standard form .

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Question

What is the length of a line with endpoints at and ?

Answer

The formula for the length of a line is very similiar to the pythagorean theorem:

Plug in our given numbers to solve:

$\sqrt{25}=\sqrt{l^2}$

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Question

What is the length of a line with endpoints at and ?

Answer

The formula for the length of a line is very similiar to the pythagorean theorem:

Plug in our given numbers to solve:

$\sqrt{25}=\sqrt{l^2}$

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Question

What is the distance between points and ?

Answer

Use the distance formula:

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

Plug in the given points:

$d=\sqrt{(12-(-2)))^2+(3-4)^2}$

$d=\sqrt{197}$

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Question

Find the distance between points and .

Answer

Use the distance formula:

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

Substitute the given points into the formula:

$d=\sqrt{(14-2)^2+(-2-(-4))^2}$

$d=\sqrt{148}$

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Question

What is the length of a line with endpoints of and ?

Answer

The distance formula is just a reworking of the Pythagorean theorem: $l=\sqrt{\Delta x^2+\Delta y^2}$

Expand that.

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

Plug in our given values.

$l=\sqrt{(7-2)^2+(9--4)^2}$

$l=\sqrt{(5)^2+(13)^2}$

$l=\sqrt{25+169}$

$l=\sqrt{194}$

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Question

What is the distance between and ?

Answer

Let $P_{1}=(1,5)$ and $P_{2} = (6,17)$ .

The distance formula is given by $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Substitute in the given points:

$d = \sqrt{(6-1)^{2}+(17-5)^{2}}=\sqrt{(5)^{2}+(12)^{2}}=\sqrt{25+144}=\sqrt{169}=13$

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Question

A line segment has endpoints at and . What is the distance of this segment?

Answer

To find the distance, we use the distance formula: $d^2=\Delta x^2+\Delta y^2$ .

Expand that:

Plug in our given values.

$\sqrt{d^2}=\sqrt{81}$

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Question

What is the midpoint of a line segment connecting the points and ?

Answer

Use the midpoint formula, $mid=(\frac{x_{1}+x_2}{2}, \frac{y_1+y_2}{2})$ , with our points and .

$mid=(\frac{-4+8}{2},\frac{6+2}{2})$

$mid=(\frac{4}{2},\frac{8}{2})=(2,4)$

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Question

What would be the midpoint of a line segment with endpoints at and ?

Answer

The midpoint of a line segment is halfway between the two $\small x$ values and halfway between the two $\small y$ values.

Mathematically, that would be the average of each coordinate: $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$ .

Plug in the values from the given points and solve.

$(\frac{5+0}{2},\frac{7+1}{2})$

$(\frac{5}{2},\frac{8}{2})$

We can simplify the fraction to give our final answer.

$(\frac{5}{2},4)$

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Question

What would be the midpoint of a line segment with endpoints at and ?

Answer

The midpoint of a line segment is halfway between the two $\small x$ values and halfway between the two $\small y$ values.

Mathematically, that would be the average of each coordinate: $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$ .

Plug in the values from the given points and solve.

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

$(\frac{10}{2},\frac{15}{2})$

Simplify the fractions to get the final answer.

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Question

If a line has a midpoint at , and the endpoints are and , what is the value of ?

Answer

The midpoint of a line segment is halfway between the two $\small x$ values and halfway between the two $\small y$ values.

Mathematically, that would be the average of the coordinates: $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$ .

Plug in the values from the given points.

$(\frac{4+0}{2},\frac{y+0}{2})=(2,5)$

Now we can solve for the missing value.

$(\frac{4}{2},\frac{y}{2})=(2,5)$

$\frac{4}{2}=2$

The $\small x$ values reduce, so both $\small x$ values equal $\small 2$ . Now we need to create a new equation to solve for the $\small y$ value.

$\frac{y}{2}=5$

Multiply both sides by to solve.

$(2)\frac{y}{2}=5(2)$

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Question

What is the midpoint of the line segment which connects and ?

Answer

To find the midpoint of a line segment, we find the average of the x and y coordinates of the endpoints. The average of two numbers is the sum of those numbers divided by . Thus, to find the x-coordinate of our midpoint, we find the average of and , and we get .

To find the y-coordinate of our midpoint, we find the average of and , which is .

Thus, our midpoint is

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Question

What is the midpoint of a line segment with endpoints and ?

Answer

The midpoint formula is this: $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$ .

Plug in the given values from our points and solve:

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

$(\frac{12+1}{2},\frac{12+1}{2})$

$(\frac{13}{2},\frac{13}{2})$

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Question

What is the midpoint of the line segment with endpoints at and ?

Answer

The midpoint formula is this: $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$ .

Plug in the given values from our points and solve:

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

$(\frac{9+6}{2},\frac{2+3}{2})$

$(\frac{15}{2},\frac{5}{2})$

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