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Two Lines


Alignments to Content Standards: 8.EE.C

Task

Consider the graph below showing two lines, L1 and L2.

Two_lines_7840026fa5565682cb78b4527fdd8a94

  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.