# Two Lines

Alignments to Content Standards: 8.EE.C

Consider the graph below showing two lines, $L1$ and $L2$. 1. Find the two corresponding linear equations.

2. Find points other than the ones given in the graph; one that lies on $L1$ but not on $L2$ and one that lies on $L2$ but not on $L1$.

## IM Commentary

In this task, we are given the graph of two lines including the coordinates of the intersection point and the coordinates of the two vertical intercepts, and are asked for the corresponding equations of the lines. It is a very straightforward task that connects graphs and equations and solutions and intersection points.

## Solution

1. The graph shows two lines, $L1$ and $L2$. In slope-intercept form, the equations are of the form $y = mx + b$. The graph shows the $y$-intercepts of both lines. Since the $y$-intercept corresponds to the point where $x=0$, the $y$-intercept of a line in slope-intercept form is equal to the value of $b$. Therefore, we have:

\begin{align} L1 : y &= m_1x+5 \\ L2 : y &= m_2x+13 \end{align}

From the graph we can also see that the point $(4, 21)$ lies on both lines. Therefore, $x = 4$, $y = 21$ is a solution to both equations. Substituting the solution into the equations, we have

\begin{align} 21 &= m_1 \cdot 4+5 \\ 21 &= m_2 \cdot 4+13. \end{align}

Solving the equations for $m_1$ and $m_2$ we find $m_1 = 4$ and $m_2 = 2$, and so our two equations are

\begin{align} y &= 4x+5 \\ y &= 2x+13 \end{align}
2. We know that a point of intersection of two graphs corresponds to the solution of the corresponding system of equations. These two lines intersect at $(4, 21)$. As lines can intersect at most once, we also know that $(4, 21)$ is the only solution to this system of equations. Therefore, to find a point on one line that is not on the other, we can take any $x$-coordinate other than 4 (or 0 to avoid the $y$-intercepts) and solve for the corresponding $y$-coordinate. For $y = 5 + 4x$ let us take $x = 2$. Then we find the corresponding $y$-value through substitution:

$$y = 5 + 4(2) = 5 + 8 = 13$$

and so $(2, 13)$ is a point on $L1$. We can verify that this point does not also lie on $L2$, again through substitution

$$y = 13 + 2(2) = 13 + 4 = 17$$

We found that at $x=2$ the $y$-coordinate for $L2$ is not $y=13$ but rather $y=17$. So, we have, in fact, found two points, $(2, 13)$ and $(2, 17)$, which lie on one line but not the other.