how to find the equation of a line with one point
Slope-Intercept Form of a Line (y = mx + b)
The slope-intercept is the most "popular" form of a straight line. Many students find this useful because of its simplicity. One can easily describe the characteristics of the straight line even without seeing its graph because the slope and y -intercept can easily be identified or read off from this form.
Slope-Intercept Form of the Equation of a Line
The linear equation written in the form
\large{y = mx + b}
is in slope-intercept form where:
m is the slope, and b is the y -intercept
![this is the slope-intercept form of a line, y=mx + b, where it is clearly labeled using arrows that the slope is denoted by the letter m and the y-intercept is denoted by letter b](https://www.chilimath.com/wp-content/uploads/2018/05/slope-intercept-form.png)
Quick notes:
- The slope m measures how steep the line is with respect to the horizontal. Given two points \left( {{x_1},{y_1}} \right) and \left( {{x_2},{y_2}} \right) found in the line, the slope is computed as
![the formula to find the slope of a line is the ratio of the difference of the y-coordinates and the difference of the x-coordinates. In an equation, we have m=(y sub2 - y sub 1)/(x sub2 - x sub1).](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10.png)
- The y-intercept b is the point where the line crosses the y-axis. Notice that in the graph below, the red dot is always found on the main vertical axis of the Cartesian plane. That is the basic characteristic of the y-intercept.
![an illustration showing the y-intercepts of two arbitrary lines in an xy-plane](https://www.chilimath.com/wp-content/uploads/2017/03/xyb.gif)
Let's go over some examples of how to write the equation of a straight line in linear form y = mx + b.
Examples of Applying the Concept of Slope-Intercept Form of a Line
Example 1: Write the equation of the line in slope-intercept form with a slope of - \,5 and a y -intercept of 3.
The needed information to write the equation of the line in the form y = mx + b are clearly given in the problem since
m = - \,5 (slope)
b = 3 (y-intercept)
Substituting in y = mx + b, we obtain
![substitute -5 for the slope and 3 for the y-intercept. This gives us y=mx+b → y=(-5)x+(3) → y=-5x+3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex1a.png)
By having a negative slope, the line is decreasing/falling from left to right, and passing through the y-axis at point \left( {0,3} \right).
![the graph of the line y=-5x+3 as shown in the coordinate plane. the line has a negative slope of m=-3 and a y-intercept at (0,3).](https://www.chilimath.com/wp-content/uploads/2017/03/ex1_g-1.gif)
Example 2: Write the equation of the line in slope-intercept form with a slope of 7 and a y -intercept of - \,4.
The slope is given as m = 7 and the y-intercept as b = - \,4. Substituting into the slope-intercept formula y = mx + b, we have
![since m=7 and b=-4, we can substitute that into the slope-intercept form of a line to get y=mx+b → y=7x-4](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex2a.png)
The slope is positive thus the line is increasing or rising from left to right, but passing through the y-axis at point \left( {0, - \,4} \right).
![this is an example of a graph of a line with positive slope and the y-intercept, also known as "b", is found on the negative portion of the y-axis. the equation of the line in slope-intercept form is y=7x-4, where the slope is equal to 7 and the y-intercept is -4.](https://www.chilimath.com/wp-content/uploads/2017/03/ex2_g-1.gif)
Example 3: Write the equation of the line in slope-intercept with a slope of 9 and passing through the point \left( {0, - \,2} \right).
This problem is slightly different from the previous two examples because the y-intercept b is not given to us up front. So our next goal is to somehow figure out what the value of b first.
However, if we examine the slope-intercept form, it should lead us to believe that we have enough information to solve for b. How?
![the slope-intercept form of a line, y=mx+b, is labeled to show how to substitute when the value of slope and a point are given.](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex3a.png)
That means m = 9, and from the given point \left( {0, - \,2} \right) we have x = 0 and y = - \,2. Let's substitute these known values into the slope-intercept formula and solve for the missing value of b.
![this shows the calculation to find the y-intercept or "b". y=mx+b → -2=9(0)+b → -2=0+b → -2=b](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex3b.png)
Now it is possible to write the slope-intercept form as
![y=9x-2](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex3c.png)
Example 4: Find the slope-intercept form of the line with a slope of - \,3 and passing through the point \left( { - 1,\,15} \right).
Again, the value of y-intercept b is not directly provided to us. But we can utilize the given slope and a point to find it.
![since we know the slope and a point on the line, we can easily compute for the value of the y-intercept](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex4a.png)
Substitute the known values into the slope-intercept formula, and then solve for the unknown value of b.
![given that m=-3 and the point is (-1,15), we have y=mx+b → 15=(-3)(-1)+b → 15=3+b → 15-3=3-3+b → 12=b](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex4b.png)
Back substitute the value of the slope and the solved value of the y-intercept into y = mx + b.
![from the last step, we found that b=12 so the equation of the line in slope-intercept form becomes y=-3x+12](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex4c.png)
Example 5: A line with the slope of - \,8 and passing through the point \left( { - \,4,\, - 1} \right).
The given slope is m = - \,8 and from the given point \left( { - \,4,\, - 1} \right), we have x = - \,4 and y = - \,1. Now, we are going to substitute the known values into the slope-intercept form of the line to solve for b.
![substituting the values, we obtain y=mx+b → -1=-8(-4)+b → -1=32+b → -1-32=32-32+b → -33=b](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex5a.png)
Since m = - \,8 and b = - \,33, the slope-intercept form of the line becomes
![y=-8x-33](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex5b.png)
Example 6: Write the slope-intercept form of the line with a slope of{3 \over 5} and through the point \left( {5,\, - 2} \right).
We have a slope here that is not an integer, i.e. the denominator is other than positive or negative one, \pm 1. In other words, we have a "true" fractional slope.
The procedure for solving this problem is very similar to examples #3, #4 and #5. But the main point of this example is to emphasize the algebraic steps required on how to solve a linear equation involving fraction.
The known values of the problem are
- Given slope:
![m=3/5](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex6a.png)
- Given point:
![x=5 and y=-2](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex6b.png)
Plug the values into y = mx + b and solve for b.
![if a line has a slope of 3/5 and passing through the point (5,-2), after substituting the values and calculating using the slope-intercept form, we find that the y-intercept is -5.](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex6c.png)
As you can see the common factors of 5 in the numerator and denominator nicely cancel each other out which greatly simplifies the process of solving for b .
Putting this together in the form y = mx + b
![y=(3/5)x-5](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex6d.png)
Example 7: Slope of{{\, - 3} \over 2} and through the point \left( { - 1,\, - 1} \right).
The given slope is m = {{\, - 3} \over 2} and from the given point\left( { - 1,\, - 1} \right), the values of x and y can easily be identified.
![(-1,-1) → x=-1 and y=-1](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex7a.png)
Now plug in the known values into the slope-intercept form y = mx + b to solve for b.
![solve for b in y=mx+b](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex7b.png)
Make sure that when you add or subtract fractions that you generate a common denominator.
![y=mx+b → -1=+b → -1 =(3/2)+b → (-2-3)/2=b therefore b=-5/2](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex7c.png)
After getting the value of b, we can now write the slope-intercept form of the line.
![y=(-3/2) multiplied to the variable x minus the fraction 5/2](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex7d.png)
Example 8: Slope of - \,6 and through the point \left( {{1 \over 2},{1 \over 3}} \right).
The slope is given as m = - \,6 and from the point, we have x = {1 \over 2} and y = {1 \over 3}.
Substitute the known values into y = mx + b. Then solve the missing value of b .
![this is a bit more challenging problem to find the y-intercept using the slope-intercept form of a line because the coordinates of the point of the line are both fractions. given that m=-6 and the point is (1/2, 1/3), after plugging in the values into y=mx+b, we find that b=10/3.](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex8a.png)
Therefore, the slope-intercept form of the line is
![y=-6x+(10/3)](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex8b.png)
Example 9: Slope of{{\,7} \over 3} and through the point \left( {{{ - \,2} \over 5},{5 \over 2}} \right).
Identifying the known values
- Given slope:
![m = 7/3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex9a.png)
- Given point:
![x = (-2)/5 and y = 5/2](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex9b.png)
The setup to find b becomes
![b=103/30](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex9c.png)
That makes the slope-intercept form of the line as
![y=mx+b → y=(7/3)x+(103/30) → y=7/3x+103/30](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex9d.png)
Example 10: A line passing through the given two points \left( {4,\,5} \right) and ( \left( {0,\,3} \right).
In this problem, we are not provided with both the slope m and y-intercept b. However, we should realize that the slope is easily calculated when two points are known using the Slope Formula.
Slope Formula
The slope, m, of a line passing through two arbitrary points \left( {{x_1},{y_1}} \right) and \left( {{x_2},{y_2}} \right) is calculated as follows…
![we divide the change in y-coordinates by the change in x-coordinates to calculate the slope of the line](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10.png)
If we let \left( {4,\,5} \right) be the first point, then \left( {0,\,3} \right) must be the second.
Labeling the components of each point should help in identifying the correct values that would be substituted into the slope formula.
![the first ordered pair (x1,y1) has the x-coordinate of 4 and a y-coordinate of 5 while the second ordered pair (x2,y2) has the x-coordinate of 0 and a y-coordinate of 3.](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10a.png)
Based on the labeling above, now we know that
![for the first point, we have x1=4, y1=5 and for the second point, we have x2=0, y2=3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10b.png)
Next, write the slope formula, plug in the known values and simplify.
![m=(y2-y1)/(x2-x1) → m=(3-5)/(0-4) → m=-2/-4 → m=1/2 or 0.5](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10d.png)
Great! We found the slope to be m = {{\,1} \over 2}\,. The only missing piece of the puzzle is to determine the y-intercept. Use the slope that we found, together with ANY of the two given points. In this exercise, I will show you that we should arrive at the same value of the y-intercept regardless which point is selected for the calculation.
Finding the y-intercept
- Using the first point \left( {4,\,5} \right).
![m=1/2 and (4,5) → x=4 and y=5](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10e.png)
![b=3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10f.png)
- Using the second point \left( {0,\,3} \right).
![m=1/2 and (0,3) → x=0 and y=3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10g.png)
![y=mx+b → 3=(1/2)(0)+b → 3=0+b → 3=b](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10h.png)
Indeed, the y-intercepts come out the same in both calculations. We can now write the linear equation in slope-intercept form.
![y=(1/2)x+3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex10i.png)
Below is the graph of the line passing through the given two points.
![the equation of the line passing through the points (0,3) and (4,5) in slope-intercept form is y=(1/2)x+3](https://www.chilimath.com/wp-content/uploads/2017/03/ex10_g.gif)
Example 11: A line passing through the given two points \left( { - \,7,\,4} \right) and \left( { - \,2,\,19} \right).
Let's solve this step by step.
- Step 1: Assign which point is the first and second, and then label its components.
![(-7,4) and (-2,19)](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex11a.png)
- Step 2: Substitute the known values into the slope formula, and simplify if necessary.
![slope is equal to 3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex11b.png)
- Step 3: Pick any of the two given points. Suppose we pick the point \left( { - \,7,\,4} \right). That means x = - \,7 and y = 4. Using the calculated value of slope in step 2, we can now find the y-intercept b.
![the y-intercept is positive 25 which can be written as b=25 or (0, 25)](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex11c.png)
- Step 4: Putting them together in y = mx + b form, since m = 3 and b = 25, we have the slope-intercept form of the line as
![the equation of the line has a slope of 3 and y-intercept of 25](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex11d.png)
- Step 5: Using a graphing utility, show that the solved linear equation in slope-intercept form passes through the two points.
![the line passing through the points (-7,4) and (-2,19) has a slope-intercept form of y=3x+25](https://www.chilimath.com/wp-content/uploads/2017/03/ex11_g.gif)
Example 12: A line passing through the given two points \left( { - \,6,\, - \,3} \right) and \left( { - \,7,\, - 1} \right).
- Find the slope
![(-6,-3) and (-7,-1)](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex11e.png)
![slope is equal to -2](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex12b.png)
- Pick any of the two given points. Suppose, we chose the second point which is
![x = -7 and y = -1](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex12c.png)
Substitute known values in the slope-intercept form y = mx + b to solve for b.
![b=-15](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex12d.png)
- Putting them together. Since m = - \,2 and b = - \,15, the slope-intercept form of the line is
![y=mx+b → y=(-2)x+(-15) → y=-2x-15](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex12e.png)
- This is the graph of the line showing that it passes both of the two points.
![the line y=-2x-15 passes through the points (-7,-1) and (-6,-3)](https://www.chilimath.com/wp-content/uploads/2017/03/ex12_g.gif)
Example 13: A line passing through the given two points \left( {5,\, - \,2} \right) and \left( { - \,2,\,5} \right).
- Determine the slope from the given two points
![the first point is (5,-2) while the second point is (-2,5)](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex13a.png)
![slope is equal to -1](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex13b.png)
- Pick any of the two given points. Let's say we chose the first one, \left( {5,\, - \,2} \right). That means x = 5, and y = - \,2. Use this information together with the value of slope to solve for the y-intercept b.
![the y-intercept is positive 3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex13c.png)
- Now, put them together. Since m = - \,1 and b = 3, the slope-intercept form of the line is
![y=-x+3](https://www.chilimath.com/wp-content/uploads/2019/03/sif-ex13d.png)
- Using a graphing utility, show that the line passes through the two given points.
![first, calculate the slope of the points (-2,5) and (5,-2). second, calculate the y-intercept using the slope that you computed previously and selecting any of the two points. once you have the slope and the y-intercept, you can write the equation of the line in the slope-intercept form which in this case is y=-x+3 as shown on the graph.](https://www.chilimath.com/wp-content/uploads/2017/03/ex13_g.gif)
Practice with Worksheets
You might also be interested in:
Types of Slopes of a Line
Slope Formula of a Line
Point-Slope Form of a Line
how to find the equation of a line with one point
Source: https://www.chilimath.com/lessons/intermediate-algebra/slope-intercept-form/
Posted by: johnsonrigh1962.blogspot.com
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