Determine an equation for a line when given two points on the line and when given the slope and one point on the line. Express these equations in slope-intercept or point-slope form and determine the slope and y-intercept of a line given an equation.
Subsection3.2.1Activities
Activity3.2.1.
Consider the graph of two lines.
(a)
Find the slope of line A.
\(\displaystyle 1\)
\(\displaystyle 2\)
\(\displaystyle \dfrac{1}{2} \)
\(\displaystyle -2\)
Answer.
B
(b)
Find the slope of line B.
1
2
\(\displaystyle \dfrac{1}{2} \)
\(\displaystyle -2\)
Answer.
B
(c)
Find the \(y\)-intercept of line A.
\(\displaystyle -2\)
\(\displaystyle -1.5\)
\(\displaystyle 1 \)
\(\displaystyle 3\)
Answer.
D
(d)
Find the \(y\)-intercept of line B.
\(\displaystyle -2\)
\(\displaystyle -1.5\)
\(\displaystyle 1 \)
\(\displaystyle 3\)
Answer.
A
(e)
What is the same about the two lines?
Answer.
Lines A and B have the same slope.
(f)
What is different about the two lines?
Answer.
Lines A and B have different \(y\)-intercepts.
Remark3.2.2.
Notice that in Activity 3.2.1 the lines have the same slope but different \(y\)-intercepts. It is not enough to just know one piece of information to determine a line, you need both a slope and a point.
Definition3.2.3.
Linear functions can be written in slope-intercept form
\begin{equation*}
f(x)=mx+b
\end{equation*}
where \(b\) is the \(y\)-intercept (or starting value) and \(m\) is the slope (or constant rate of change).
Activity3.2.4.
Write the equation of each line in slope-intercept form.
(a)
\(\displaystyle y= -3x+1\)
\(\displaystyle y= -x+3\)
\(\displaystyle y= -\dfrac{1}{3}x+1\)
\(\displaystyle y= -\dfrac{1}{3}x+3\)
Answer.
C
(b)
The slope is \(4\) and the \(y\)-intercept is \((0,-3)\text{.}\)
\(\displaystyle f(x)= 4x-3\)
\(\displaystyle f(x)= 3x-4\)
\(\displaystyle f(x)= -4x+3\)
\(\displaystyle f(x)= 4x+3\)
Answer.
A
(c)
Two points on the line are \((0,1)\) and \((2,4)\text{.}\)
\(\displaystyle y= 2x+1\)
\(\displaystyle y= -\dfrac{3}{2}x+4\)
\(\displaystyle y= \dfrac{3}{2}x+1\)
\(\displaystyle y= \dfrac{3}{2}x+4\)
Answer.
C
(d)
\(x\)
\(f(x)\)
\(-2\)
\(-8\)
\(0\)
\(-2\)
\(1\)
\(1\)
\(4\)
\(10\)
\(\displaystyle f(x)= -3x-2\)
\(\displaystyle f(x)= -\dfrac{1}{3}x-2\)
\(\displaystyle f(x)= 3x+1\)
\(\displaystyle f(x)= 3x-2\)
Answer.
D
Activity3.2.5.
Let’s try to write the equation of a line given two points that don’t include the \(y\)-intercept.
(a)
Plot the points \((2,1)\) and \((-3,4)\text{.}\)
Answer.
(b)
Find the slope of the line joining the points.
\(\displaystyle -\dfrac{5}{3}\)
\(\displaystyle -\dfrac{3}{5}\)
\(\displaystyle \dfrac{3}{5}\)
\(\displaystyle -3\)
Answer.
B
(c)
When you draw a line connecting the two points, it’s often hard to draw an accurate enough graph to determine the \(y\)-intercept of the line exactly. Let’s use the slope-intercept form and one of the given points to solve for the \(y\)-intercept. Try using the slope and one of the points on the line to solve the equation \(y=mx+b\) for \(b\text{.}\)
\(\displaystyle 2\)
\(\displaystyle \dfrac{11}{5}\)
\(\displaystyle \dfrac{5}{2}\)
\(\displaystyle 3\)
Answer.
B
(d)
Write the equation of the line in slope-intercept form.
Answer.
\(y=-\dfrac{3}{5}x+\dfrac{11}{5}\)
Remark3.2.6.
It would be nice if there was another form of the equation of a line that works for any points and does not require the \(y\)-intercept.
Definition3.2.7.
Linear functions can be written in point-slope form
\begin{equation*}
y-y_0=m(x-x_0)
\end{equation*}
where \((x_0, y_0)\) is any point on the line and \(m\) is the slope.
Activity3.2.8.
Write an equation of each line in point-slope form.
(a)
\(\displaystyle y= \dfrac{1}{3}x+\dfrac{2}{3}\)
\(\displaystyle y-1= 3(x-1)\)
\(\displaystyle y-1= \dfrac{1}{3}(x-1)\)
\(\displaystyle y+2= \dfrac{1}{3}(x+2)\)
\(\displaystyle y= \dfrac{1}{3}(x+2)\)
Answer.
C or E
(b)
The slope is \(4\) and \((-1,-7)\) is a point on the line.
\(\displaystyle y+7= 4(x+1)\)
\(\displaystyle y-7= 4(x-1)\)
\(\displaystyle y+1= 4(x+7)\)
\(\displaystyle y-4= 7(x-1)\)
Answer.
A
(c)
Two points on the line are \((1,0)\) and \((2,-4)\text{.}\)
\(\displaystyle y= -4x+1\)
\(\displaystyle y-0=-2(x-1) \)
\(\displaystyle y+4=-4(x-2)\)
\(\displaystyle y+4=-3(x-2)\)
Answer.
C
(d)
\(x\)
\(f(x)\)
\(-2\)
\(-8\)
\(1\)
\(1\)
\(4\)
\(10\)
\(\displaystyle y+8= 3(x-2)\)
\(\displaystyle y-1= -\dfrac{1}{3}(x-1)\)
\(\displaystyle y+8= -\dfrac{1}{3}(x+2)\)
\(\displaystyle y-10= 3(x-4)\)
Answer.
D
Activity3.2.9.
Consider again the two points from Activity 3.2.5, \((2,1)\) and \((-3,4)\text{.}\)
(a)
Use point-slope form to find an equation of the line.
\(\displaystyle y=-\dfrac{3}{5}x+\dfrac{11}{5}\)
\(\displaystyle y-1=-\dfrac{3}{5}(x-2)\)
\(\displaystyle y-4 =-\dfrac{3}{5}(x+3)\)
\(\displaystyle y-2=-\dfrac{3}{5}(x-1) \)
Answer.
B or C
(b)
Solve the point-slope form of the equation for \(y\) to rewrite the equation in slope-intercept form. Identify the slope and intercept of the line.
Answer.
The slope-intercept form is: \(y=-\dfrac{3}{5}x+\dfrac{11}{5}\text{.}\) The slope is \(-\dfrac{3}{5}\) and the \(y\)-intercept is \(\dfrac{11}{5}\text{.}\)
Remark3.2.10.
Notice that it was possible to use either point to find an equation of the line in point-slope form. But, when rewritten in slope-intercept form the equation is unique.
Activity3.2.11.
For each of the following lines, determine which form (point-slope or slope-intercept) would be "easier" and why. Then, write the equation of each line.
(a)
Answer.
Slope-intercept: \(y=\dfrac{3}{4}x+2\)
(b)
The slope is \(-\dfrac{1}{2}\) and \((1,-3)\) is a point on the line.
Answer.
Point-slope: \(y+3=-\dfrac{1}{2}(x-1)\)
(c)
Two points on the line are \((0,3)\) and \((2,0)\text{.}\)
Answer.
Slope-intercept: \(y=-\dfrac{3}{2}x+3\)
Remark3.2.12.
It is always possible to use both forms to write the equation of a line and they are both valid. Although, sometimes the given information lends itself to make one form easier.
Activity3.2.13.
Write the equation of each line.
(a)
The slope is \(0\) and \((-1,-7)\) is a point on the line.
\(\displaystyle y=-7\)
\(\displaystyle y=7x\)
\(\displaystyle y=-x\)
\(\displaystyle x=-1\)
Answer.
A
(b)
Two points on the line are \((3,0)\) and \((3,5)\text{.}\)
\(\displaystyle y=3x+3\)
\(\displaystyle y=3x+5 \)
\(\displaystyle x=3\)
\(\displaystyle y=3\)
Answer.
C
(c)
\(\displaystyle x=-2 \)
\(\displaystyle y-2=x\)
\(\displaystyle y=-2x-2\)
\(\displaystyle y=-2\)
Answer.
D
Definition3.2.14.
A horizontal line has a slope of zero and has the form \(y=k\) where \(k\) is a constant. A vertical line has an undefined slope and has the form \(x=h\) where \(h\) is a constant.
Definition3.2.15.
The equation of a line can also be written in standard form. Standard form looks like \(Ax+By=C\text{.}\)
Remark3.2.16.
It is possible to rearrange a line written in standard form to slope-intercept form by solving for \(y\text{.}\)
