Algebra - How to find the equation of a line

Determine the equation for the line, in slope intercept form, given a slope of and that it passes through the point .

Explanation

To determine the equation for the line given only a point on the line and its slope, we can use point-slope form, which is given by the following:

$y-y_{1}=m(x-x_{1})$ , where is the slope of the line and $(x_{1}, y_{1})$ is the point on the line.

Using the formula, we get

, which simplified becomes .

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

Explanation

Write the slope-intercept form for linear equations.

The is the slope of the equation, and is the y-intercept.

Substitute the values into the equation.

The equation of the line is:

Write the slope-intercept form of the equation of the line described.

Passes through the point , perpendicular to $y=-\frac{2}5{}x-3$ .

$y=\frac{5}{2}x-\frac{13}{2}$

$y=-\frac{2}{5}x+\frac{11}{5}$

$y=-\frac{2}5{}x+\frac{13}{5}$

$y=-\frac{5}{2}x+\frac{17}{2}$

$y=-\frac{1}{2}x+\frac{11}{5}$

Explanation

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

A line that is perpendicular has a slope that is the opposite reciprocal of the given line.

Slope of perpendicular line:

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

Using the point slope formula,

$y-y_{1}=m(x-x_{1})$

where $x_{1}=3: :and: :y_{1}=1$

we get the following equation.

$y-1=\frac{5}{2}(x-3)$

$y=\frac{5}{2}x-\frac{15}{2}+1$

$y=\frac{5}{2}x-\frac{13}{2}$

Find the equation of the line with slope 4 running through the point (-1,-5).

Explanation

To solve this problem, we need to remember point-slope formula:

$y-y_{1}=m(x-x_{1})$

Then we plug in m=4 and (x1,y1)=(-1,-5) and solve:

Find the equation of the line given two points: and

$y=-\frac{2}{3}x+9\frac{2}{3}$

$y=-x+9\frac{2}{3}$

$y=-\frac{4}{3}x+7\frac{5}{8}$

$y=-\frac{2}{5}x+8\frac{1}{3}$

Explanation

Write the slope intercept form. The equation will be in this form.

Write the slope formula.

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

Let and .

Substitute the points into the slope formula.

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{11-9}{-2-1} = \frac{2}{-3}$

The slope is: $-\frac{2}{3}$

Use the slope and a given point, substitute them into the slope intercept form to find the y-intercept.

$9=(-\frac{2}{3})(1)+b$

Solve for the y-intercept.

$9=-\frac{2}{3}+b$

Add two-thirds on both sides.

$9+ \frac{2}{3}=-\frac{2}{3}+b+ \frac{2}{3}$

Simplify both sides.

$b=9\frac{2}{3}$

With the slope and y-intercept known, write the formula.

The answer is: $y=-\frac{2}{3}x+9\frac{2}{3}$

$\small (4,2) (5,6)$ $\small y-intercept = 4$

Using the data provided, find the equation of the line.

$\small y = {4}x + 4$

$\small y = 4x + 5$

$\small y = 3x -1$

$\small y = 6x + 21$

$\small y = \frac{3}{2}x + 5$

Explanation

The equation for a line is always $\small y = mx + b$ , where $\small m = slope$ and $\small b = y-intercept$ .

Given two data points, we are able to find the slope, m, using the formula

$\small \frac{y_{1}-y_{2}}{x_{1}-x_{2}}$ .

Using the data points provided, our formula will be:

$\small \frac{6-2}{5-4}$ , which gives us $\small \frac{4}{1}=4$ or , $\small m = 4$ .

Our y-intercept is given.

Thus our equation for the line containing these points and that y-intercept is $\small y = {4}x + 4$ .

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

y = 3x – 5

y = 3x + 5

y = 5x – 3

y = –5x + 3

y = (3/5)x + 2

Explanation

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

Write an equation in the form for the line that fits the following points:

(4,3), (6,6), (10,12)

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

$y=-\frac{3}{2}x-6$

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

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

Explanation

The equation of a line is written in the following format:

The first step, then, is to find the slope, .

is equal to the change in divided by the change in .

So,

$\frac{6-3}{6-4}=\frac{3}{2}$

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

Next step is to find . We can find values for and from any one of the given points, plug them in, and solve for .

Let's use (4,3)

So,

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

Then we just fill in our value for , and we have as a function of .

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

Find the equation of the line that contains the points (4, 5) and (-2, -1).

Explanation

When finding the equation of a line given two points, we must first find the slope of the line. To find the slope of a line in slope-intercept form

where m is the slope, we calculate

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

where and are the points. Given the points, we can substitue. We get

$m = \frac{-1 - 5}{-2 - 4}$

$m = \frac{-6}{-6}$

Now that we know the slope, we can substitute the slope into the slope-intercept form

All we need to do now if find the value of b. To do that, we will substitute one of the points into the equation. Let's use (4, 5). So,

$5 = 1 \cdot 4 + b$

So if we substitute the value of b into the slope-intercept form with the slope included, we get

which can also be written as

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

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

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

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

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

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

Explanation

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

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

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

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

How to find the equation of a line

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