How to find the equation of a line - Algebra 1

Card 0 of 20

Question

Given two points, (5, –8) (–2, 6), what is the equation of the line containing them both?

Answer

First, you should plug the given points, (5, –8) (–2, 6), into the slope formula to find the slope of the line.

$slope=\frac{y_2-y_1}{x_2-x_1}$

$slope=\frac{-8-6}{5-(-2)}=\frac{-8-6}{5+2}=\frac{-14}{7}=-2$

Then, plug the slope into the slope formula, y = mx + b, where m is the slope.

y = –2x + b

Plug in either one of the given points, (5, –8) or (–2, 6), into the equation to find the y-intercept (b).

6 = –2(–2) + b

6 = 4 + b

2 = b

Plug in both the slope and the y-intercept into slope intercept form.

y = –2x + 2

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Question

What is the equation of the line connecting the points and ?

Answer

To find the equation of this line, we need to know its slope and y-intercept. Let's find the slope first using our general slope formula.

$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$

The points are and . In this case, our points are (–3,0) and (2,5). Therefore, we can calculate the slope as the following:

$m=\frac{5-0}{2-(-3)}=\frac{5}{5}=1$

Our slope is 1, so plug that into the equation of the line:

$y=mx+b ,,, \rightarrow ,,,y=(1)x+b ,,, \rightarrow ,,, y=x+b$

We still need to find b, the y-intercept. To find this, we pick one of our points (either (–3,0) or (2,5)) and plug it into our equation. We'll use (–3,0).

Solve for b.

The equation is therefore written as .

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Question

What is the equation of a line with slope of 3 and a y-intercept of –5?

Answer

These lines are written in the form y = mx + b, where m is the slope and b is the y-intercept. We know from the question that our slope is 3 and our y-intercept is –5, so plugging these values in we get the equation of our line to be y = 3x – 5.

m = 3 and b = –5

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Question

Given two points $\left ( 1,5 \right )$ and $\left ( 0,2 \right )$ , find the equation of a line that passes through the point $\left ( -3,-5 \right )$ and is parallel to the line passing through points and .

Answer

The slope of the line passing through points and can be computed as follows:

$Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-2}{1-0}=3$

Now, the new line, since it is parallel, will have the same slope. To find the equation of this new line, we use point-slope form:

$y-y_{1}=m(x-x_{1})$ , where is the slope and $\left ( x_{1},y_{1} \right )$ is the point the line passes through.

After rearranging, this becomes

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Question

A line contains the points (8, 3) and (-4, 9). What is the equation of the line?

Answer

is the slope-intercept form of the equation of a line.

Slope is equal to $\frac{rise}{run}$ between points, or $\frac{6}{-12}$ .

So $m=-\frac{1}{2}$ .

At point (8, 3 ) the equation becomes

$3=-\frac{1}{2}\cdot 8+b\rightarrow 3=-4+b$

So

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Question

Find the equation, in form, of the line that contains the points and .

Answer

When finding the equation of a line from some of its points, it's easiest to first find the line's slope, or .

To find slope, divide the difference in values by the difference in values. This gives us divided by , or $-\frac{1}{2}$ .

Next, we just need to find , which is the line's -intercept. By plugging one of the points into the equation $y = -\frac{1}{2}x+b$ , we obtain a value of 11 and a final equation of

$y=-\frac{1}{2}x+11$

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Question

Find the equation for the line goes through the two points below.

Answer

Let $P_{1} = (1, ; -2) ; ; and ; ; P_{2} = (-3, ; 6)$ .

First, calculate the slope between the two points.

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{6 - (-2)}{-3 - 1} = \frac{8}{-4} = -2$

Next, use the slope-intercept form to calculate the intercept. We are able to plug in our value for the slope, as well the the values for .

Using slope-intercept form, where we know and , we can see that the equation for this line is .

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Question

What is the equation of a line with a slope of and a -intercept of ?

Answer

When a line is in the format, the is its slope and the is its -intercept. In this case, the equation with a slope of and a -intercept of is .

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Question

What is the equation of a straight line that connects the points indicated in the table?

Answer

We can find the equation of th line in slope-intercept form by finding and .

First, calculate the slope, , for any two points. We will use the first two.

$m=\frac{y_{2}-y_1}{x_2-x_1}=\frac{7-4}{1-0}=\frac{3}{1}=3$

Next, using the slope and any point on the line, calculate the y-intercept, . We will use the first point.

The correct equation in slope-intercept form is .

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Question

$\textup{What is the equation of the line in the }xy\textup{-coordinate plane that}$

$\textup{passes through the points }\left ( -6,4\right )\textup{and}\left ( -2,2\right )?$

Answer

$\textup{Equation of a line is }y=mx+b \textup{, where }m\textup{ is the slope and }b\textup{ is the }y\textup{-intercept.}$

$\textup{Find slope} = \frac{\Delta y}{\Delta x}=\frac{2-4}{-2--6}=\frac{-2}{4}=\frac{-1}{2}$

$\textup{Use slope to find }y\textup{-intercept:} \frac{-1}{2}= \frac{b-2}{0--2};;;;\frac{-1}{2}=\frac{b-2}{2}$

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

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Question

In 1990, the value of a share of stock in General Vortex was $27.17. In 2000, the value was $48.93. If the value of the stock rose at a generally linear rate between those two years, which of the following equations most closely models the price of the stock, , as a function of the year, ?

Answer

We can treat the price of the stock as the value and the year as the value, making any points take the form , or . This question is asking for the line that includes points and .

To find the equation, first, we need the slope.

$m= \frac{48.93-27.17}{2000-1990}= \frac{21.76}{10}=2.176$

Now use the point-slope formula with this slope and either point (we will choose the second).

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Question

Which of these lines has a slope of 5 and a -intercept of 6?

Answer

When an equation is in the form, the indicates its slope while the indicates its -intercept. In this case, we are looking for a line with a of 5 and a of 6, or .

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Question

Which of these lines has a slope of and a -intercept of ?

Answer

When a line is in the form, the is its slope and the is its -intercept. Thus, the only line with a slope of and a -intercept of is

.

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Question

What is the equation of a line with a slope of 3 that runs through the point (4,9)?

Answer

You can find the equation by plugging in all of the information to the formula.

The slope (or ) is 3. So, the equation is now .

You are also given a point on the line: (4,9), which you can plug into the equation:

Solve for to get .

Now that you have the and , you can determine that the equation of the line is .

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Question

What is the slope and y-intercept of ?

Answer

The easiest way to determine the slope and y-intercept of a line is by rearranging its equation to the form. In this form, the slope is the and the y-intercept is the .

Rearranging

gives you

which has an of 2 and a of 6.

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Question

Find the equation of the line, in form, that contains the points , , and .

Answer

When finding the equation of a line given two or more points, the first step is to find the slope of that line. We can use the slope equation, $\frac{y_{2}-y_{_{1}}}{x_{2}-x_{1}}$ . Any combination of the three points can be used, but let's consider the first two points, and .

$\frac{12-6}{11-1} = \frac{3}{5}$

So $\frac{3}{5}$ is our slope.

Now, we have the half-finished equation

$y = \frac{3}{5}x + b$

and we can complete it by plugging in the and values of any point. Let's use .

Solving

$6 = \frac{3}{5}(1)+b$

for gives us

$b = 6 - \frac{3}{5}$

so

$b = 5\frac{2}{5} = \frac{27}{5}$

We now have our completed equation:

$y = \frac{3}{5}x + \frac{27}{5}$

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Question

Find the domain of:

$\sqrt{2x-5}$

Answer

The expression under the radical must be $\geq 0$ . Hence

$\sqrt{2x-5} \geq 0$

Solving for , we get

$x\geq \frac{5}{2}$

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Question

What is the equation of the line passing through the points (1,2) and (3,1) ?

Answer

First find the slope of the 2 points:

$\frac{1-2}{3-1}=\frac{-1}{2}$

Then use the slope and one of the points to find the y-intercept:

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

$\frac{4}{2}+\frac{1}{2}=-\frac{1}{2}+\frac{1}{2}+b$

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

So the final equation is $y=-\frac{1}{2}x+\frac{5}{2}$

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Question

We have two points: and .

If these two points are connected by a straight line, find the equation describing this straight line.

Answer

We need to find the equation of the line in slope-intercept form.

In this formula, is equal to the slope and is equal to the y-intercept.

To find this equation, first, we need to find the slope by using the formula for the slope between two point.

$m=\frac{y_2-y_1}{x_2-x_1}$

In the formula, the points are and . In our case, the points are and . Using our values allows us to solve for the slope.

$m=\frac{9-(-1)}{4-(-3)}=\frac{10}{7}$

We can replace the variable with our new slope.

$y=\frac{10}{7}x+b$

Next, we need to find the y-intercept. To find this intercept, we can pick one of our given points and use it in the formula.

$-1=\frac{10}{7}(-3)+b$

Solve for .

$-1=\frac{-30}{7}+b$

$b=-1+\frac{30}{7}$

$b=\frac{23}{7}$

Now, the final equation connecting the two points can be written using the new value for the y-intercept.

$y=\frac{10}{7}x+\frac{23}{7}$

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Question

Give, in slope-intercept form, the equation of a line through the points and .

Answer

First, use the slope formula to find the slope, setting $x_{1} = -2, y_{1} = 5,x_{2} = -3, y_{2} = 8$ .

$m = \frac{y _{2} - y_{1}}{x _{2} - x_{1}} = \frac{8 -5 }{-3-(-2)} = \frac{3 }{-1} = -3$

We can write the equation in slope-intercept form as

.

Replace :

We can find by substituting for using either point - we will choose :

The equation is .

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