One Point, One Slope, One Line

One Point, One Slope, One Line exercise

• Find the equation for the corresponding line given a point and a slope.

  Hints

  The $y$-intercept for a line always has the form $(0,b)$. The slope, $m$, is always rise over run. The equation for a line in slope intercept form is $y=mx+b$.

  Consider a line going through the point $(0,2)$, with a slope of $3$.

  • We can use this point and slope to graph the line.
  • $(0,2)$ is the $y$-intercept. To graph it, we put a point on the $y$-axis at the $y$-value of $2$.
  • The slope of $3$ means a rise of $3$ and a run of $1$. So we go $1$ unit horizontally then $3$ units vertically to graph the next point, $(1,5)$.
  • We draw the line through these points to complete our graph as shown.
  • The equation for the line is $y=3x+2$.

  Solution

  Here is a table showing the relationship between the given information, and the graph and equation for each resulting line:

  $ \begin{array}{c|c|c|c} \text{point} & y\text{-int.} & \text{slope} & \text{equation}\\ \hline &&&\\ (0,8) & 8 & 2=\frac21 & y=2x+8\\ &&&\\ (0,-5) & -5 & -\frac12 & y=-\frac12 x - 5 \end{array} $

  To graph the line $y=2x+8$:

  • Graph the $y$-intercept, $(0,8)$.
  • Use the slope of $2$ to run $1$ and rise $2$.
  • Graph this new point, $(1,10)$.
  • Run and rise again to graph $(2,12)$.
  • Graph the line through these points.

  To graph the line $y=-\frac12 x-5$:

  • Graph the $y$-intercept, $(0,-5)$.
  • Use the slope of $-\frac12$ to run $-2$ and rise $1$.
  • Graph this new point, $(-2,-4)$.
  • Run and rise again to graph $(-4,-3)$.
  • Graph the line through these points.

• Given a point and a slope, graph the resulting line.

  Hints

  The lines below are given in slope-intercept form. The slope and the $y$-intercept are identified.

  $ \begin{array}{c|c|c|c} y=mx+b & m & b & 3 ~\text{points}\\ \hline &&&\\ y=2x+3 & 2=\frac21 & 3 & (0,3)~(1,5)~(2,7)\\ &&&\\ y=-x+6 & -1=-\frac11 & 6 & (0,6)~(1,5)~(2,4) \end{array} $

  Given the line $y=x-12$:

  • The slope is $1$.
  • The rise over run is $\frac11$.
  • From a given point, we go $1$ unit horizontally then $1$ unit vertically to get to another point on the line.

  Given the point $(0,1)$ and a slope of $5$:

  • The equation is $y=5x+1$.
  • The points $(0,1)$, $(1,6)$ and $(2,11)$ are on the line.

  Solution

  We know the line has a slope of $1$ and goes through the point $(0,2)$. This means:

  • The y-intercept is $2$.
  • The rise over run is $\frac11$.
  • From the y-intercept, we go $1$ unit horizontally and $1$ unit vertically to get to another point on the graph.
  • We can repeat that movement to get additional points.
  • The points $(0,2)$, $(1,3)$, and $(2,4)$ are on the graph of the line.

• Write the equation of the resulting line given a point and a slope.

  Hints

  Slope intercept form for a line is $y=mx+b$. The slope is $m$ and the $y$-intercept is $b$.

  Here are some examples of points and slopes, and the equations for the resulting lines:

  $ \begin{array}{l|l|l} \text{point} & \text{slope} & \text{equation}\\ \hline &&\\ (0,-15) & m=\frac89 & y=(8/9)x-15\\ &&\\ (0,0) & m=-12 & y=-12x\\ &&\\ (0,-\frac12) & m=-\frac18 & y=(-1/8)x-1/2 \end{array} $

  Solution

  $\begin{array}{l|l|l} \text{point} & \text{slope} & \text{equation}\\ \hline &&\\ (0,-15) & m=4 & y=4x-15\\ &&\\ (0,0) & m=\frac12 & y=(1/2)x\\ &&\\ (0,-101) & m=\frac47 & y=(4/7)x-101\\ &&\\ (0,-1) & m=-1 & y=-x-1\\ &&\\ (0,-4) & m=6 & y=6x-4 \end{array}$

• Determine a line from a given point and slope.

  Hints

  Here are a few examples of points and slopes, and the equations for the resulting lines:

  $\begin{array}{l|l|l} \text{point} & \text{slope} & \text{equation}\\ \hline &&\\ (0,0) & m=3 & y=3x\\ &&\\ (0,14) & m=0 & y=14\\ &&\\ (0,12) & m=-\frac17 & y=-\frac17 x + 12\\ \end{array}$

  Solution

  $\begin{array}{l|l|l} \text{point} & \text{slope} & \text{equation}\\ \hline &&\\ (0,\frac12) & m=2 & y=2x+\frac12\\ &&\\ (0,-21) & m=\frac49 & y=\frac49 x -21\\ &&\\ (0,-8) & m=0 & y=-8\\ &&\\ (0,0) & m=0 & y=0\\ &&\\ (0,1) & m=1 & y=x+1\\ \end{array}$

• Determining which points lie on which lines.

  Hints

  Given the equation $y=mx+b$:

  • The slope is $m$.
  • The $y$-intercept is $b$.
  • The point $(0,b)$ is on the graph of the line.

  Given the line $y=-\frac16 x +10$:

  • The $y$-intercept is $10$.
  • The point $(0,10)$ is on the line.
  • The slope is $-\frac16$.
  • The rise is $-1$ and the run is $6$.
  • From the point $(0,10)$, we can run $6$ and "rise" $-1$.
  • This gives us the point $(6,9)$, which is also on the line.

  Suppose we are given the line $y=\frac13 x + 11$ and a point $(3,12)$ which we don't know is on the line or not.

  We can substitute the point in for $x$ and $y$ in the equation. If the result is true, the point is on the line.

  $\begin{array}{l} 12=\frac13(3)+11\\ 12=1+11\\ 12=12 \end{array} $

  That's true, so the point $(3,12)$ is on the graph of the line $y=\frac13 x +11$.

  Solution

  $ \begin{array}{c|c|c|c} y=mx+b & \text{slope} & \text{y-int.} & \text{points}\\ \hline &&&\\ y=2x-7 & 2=\frac21 & -7 & (0,-7)\\ &&&\\ y=7x & 7=\frac71 & 0 & (0,0)\\ &&&\\ y=\frac17 x +7 & \frac17 & 7 & (0,7)\\ &&&\\ y=-\frac17 x + 1 & -\frac17 & 1 & (0,1)~(7,0)\\ &&&\\ y=7x-14 & 7=\frac71 & -14 & (0,-14) \end{array} $

• Prove that given a slope and point, there's only one line with that slope passing through that point.

  Hints

  Plugging the point $(s,t)$ into $y=mx+b$ gives us $t=ms+b$.

  If $t=ms+b_1$ and $t=ms+b_2$ then $ms+b_1=ms+b_2$.

  If $ms+b_1=ms+b_2$, then $b_1=b_2$.

  Solution

  Let’s say that we have two lines, $y=mx+b_1$ and $y=mx+b_2$, with the same slope $m$, and that both of these lines pass through the point $(s,t)$. Then we know that if we plug the point $(s,t)$ into $y=mx+b_1$ and $y=mx+b_2$, then both of our equations will be true. So, plugging $(s,t)$ in, we have that $t=ms+b_1$ and $t=ms+b_2$. Since the left-hand sides of both of these equations are both equal to $t$, we can set them equal to each other to get $ms+b_1=ms+b_2$. Using opposite operations, we subtract $ms$ from both sides of the equation to get $b_1=b_2$. But if $b_1=b_2$, then the two lines, $y=mx+b_1$ and $y=mx+b_2$, must be the same line after all! So, indeed, there is really only one line with slope $m$ which passes through the point $(s,t)$.