A linear equation is graphed on the coordinate plane below.

The graph shows a coordinate grid with a straight line that passes through the points (negative 9, negative 5), (negative 6, negative 4) (negative 3, negative 3), (0, negative 2), (3, negative 1), (6, 0), and (9, 1).

When graphed on the same coordinate plane, which equation results in a system of equations with exactly one solution?

A.

B.

C.

D.

1 answer

To determine a linear equation that results in a system with exactly one solution when graphed alongside the given line, we need to ensure that the new line intersects the existing line at a single point. This requires that the new line has a different slope than that of the original line.

  1. Identify the slope of the original line: We can find the slope (m) of the line using two of the given points. The slope formula is:
    \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

    Let's use the points (0, -2) and (3, -1): \[ m = \frac{-1 - (-2)}{3 - 0} = \frac{1}{3} \]

    Therefore, the slope of the original line is \(\frac{1}{3}\).

  2. Find an equation with a different slope: To ensure that we get exactly one solution, we need to choose an equation with a different slope. Without specific equations provided, choose any line that has a slope not equal to \(\frac{1}{3}\).

For example:

  • If you choose a line with a slope of 1, the equation could be \(y = x + b\) for some value of b that ensures it intersects the original line.
  • If you choose a line with a slope of -1, the equation could be \(y = -x + c\).
  1. Verify the choice: Make sure to plot the new line on the graph. If it intersects at exactly one point, then that is the desired equation.

Conclusion:

Without the specific options A, B, C, and D, ensure that the new equation has a slope different from \(\frac{1}{3}\). For the situation described, lines with slopes such as 1, -1, or any slope other than \(\frac{1}{3}\) will yield a system with exactly one solution. If certain options were provided, find one with a slope that differs from \(\frac{1}{3}\).