Question

svlc algebra 1a - standard (15260)
slope of a line

vera wants to graph a line that passes through (0, 2) and has a slope of \\(\frac{2}{3}\\). which points could vera use to graph the line? choose three correct answers.

\\((-2, -3)\\) \\((3, 4)\\)
\\((-3, 0)\\) \\((2, 5)\\)
\\((6, 6)\\)

Explanation:

The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \), where \( m \) is the slope, and \( (x_1, y_1) \) and \( (x_2, y_2) \) are two points on the line. We know the line passes through \( (0, 2) \) and has a slope of \( \frac{2}{3} \). So we can check each point by plugging into the slope formula with \( (0, 2) \).

Step 1: Check \((3, 4)\)

Let \( (x_1, y_1) = (0, 2) \) and \( (x_2, y_2) = (3, 4) \).
\( m = \frac{4 - 2}{3 - 0} = \frac{2}{3} \). This matches the slope.

Step 2: Check \((-3, 0)\)

Let \( (x_1, y_1) = (0, 2) \) and \( (x_2, y_2) = (-3, 0) \).
\( m = \frac{0 - 2}{-3 - 0} = \frac{-2}{-3} = \frac{2}{3} \). This matches the slope.

Step 3: Check \((6, 6)\)

Let \( (x_1, y_1) = (0, 2) \) and \( (x_2, y_2) = (6, 6) \).
\( m = \frac{6 - 2}{6 - 0} = \frac{4}{6} = \frac{2}{3} \). This matches the slope.

Step 4: Check \((-2, -3)\)

Let \( (x_1, y_1) = (0, 2) \) and \( (x_2, y_2) = (-2, -3) \).
\( m = \frac{-3 - 2}{-2 - 0} = \frac{-5}{-2} = \frac{5}{2}
eq \frac{2}{3} \).

Step 5: Check \((2, 5)\)

Let \( (x_1, y_1) = (0, 2) \) and \( (x_2, y_2) = (2, 5) \).
\( m = \frac{5 - 2}{2 - 0} = \frac{3}{2}
eq \frac{2}{3} \).

Answer:

The correct points are:

• \((3, 4)\)
• \((-3, 0)\)
• \((6, 6)\)