Functions and Lines - Math

Card 0 of 232

Question

Find the equation of a line perpendicular to

Answer

Since a perpendicular line has a slope that is the negative reciprocal of the original line, the new slope is $\frac{1}{3}$ . There is only one answer with the correct slope.

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Question

Which of the following lines will be parallel to $y=\frac{3}{4}x+12$ ?

Answer

Two lines are parallel if they have the same slope. When a line is in standard form, the is the slope.

For the given line $y=\frac{3}{4}x+12$ , the slope will be $\frac{3}{4}$ . Only one other line has a slope of $\frac{3}{4}$ : $y=\frac{3}{4}x-5$

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Question

Are the following lines parallel?

$y = \frac{1}{2}x - 5$

Answer

By definition, two lines are parallel if they have the same slope. Notice that since we are given the lines in the format, and our slope is given by , it is clear that the slopes are not the same in this case, and thus the lines are not parallel.

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Question

Which of the following equations are parallel to ?

Answer

For one equation to be parallel to another, the only requirement is that they must have the same slope. In order to figure out which answer choice is parallel to the given equation, you must first find the slope of the equation:

$y=\frac{2}{5}x-3$

From the simplified equation, you can see that the slope is $\frac{2}{5}$ .

The answer choice that has the same slope is $y=\frac{2}{5}x+2$ .

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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

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

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 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 equation, in slope-intercept form, of the perpendicular bisector of the line segment that connects the points $\small (2,2)$ and $\small (8,6)$ ?

Answer

First, calculate the slope of the line segment between the given points.

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{8-2}=\frac{4}{6}=\frac{2}{3}$

We want a line that is perpendicular to this segment and passes through its midpoint. The slope of a perpendicular line is the negative inverse. The slope of the perpendicular bisector will be $m=-\frac{3}{2}$ .

Next, we need to find the midpoint of the segment, using the midpoint formula.

$mid=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{x})=(\frac{2+8}{2},\frac{2+6}{2})=(5,4)$

Using the midpoint and the slope, we can solve for the value of the y-intercept.

$4=(-\frac{3}{2})(5)+b$

$4=\frac{8}{2}=-\frac{15}{2}+b$

$\frac{23}{2}=b$

Using this value, we can write the equation for the perpendicular bisector in slope-intercept form.

$y=-\frac{3}{2}x+\frac{23}{2}$

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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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