Vera wants to graph a line that passes through (0, 2) and has a slope of 2/3. Which points could Vera use to graph the line? Select three options.

A. (-3, 0)
B. (-2, -3)
C. (2, 5)
D. (3, 4)
E. (6, 6)



Answer :

To determine which points Vera can use to graph the line that passes through [tex]\((0,2)\)[/tex] and has a slope of [tex]\(\frac{2}{3}\)[/tex], we start by using the point-slope form of a linear equation:

[tex]\[ y = mx + b \][/tex]

Where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept. Given the slope [tex]\(m = \frac{2}{3}\)[/tex] and the point [tex]\((0,2)\)[/tex], we can find the y-intercept [tex]\(b\)[/tex].

First, we use the given point [tex]\((0,2)\)[/tex]:
[tex]\[ y = mx + b \][/tex]
[tex]\[ 2 = \left( \frac{2}{3} \right) \cdot 0 + b \][/tex]
[tex]\[ 2 = b \][/tex]

So, the equation of the line is:
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]

Next, we need to check which of the given points satisfy this equation.

Check (-3, 0):
[tex]\[ y = \frac{2}{3}(-3) + 2 \][/tex]
[tex]\[ y = -2 + 2 \][/tex]
[tex]\[ y = 0 \][/tex]

Since the coordinates are [tex]\((-3,0)\)[/tex], this point lies on the line.

Check (-2, -3):
[tex]\[ y = \frac{2}{3}(-2) + 2 \][/tex]
[tex]\[ y = -\frac{4}{3} + 2 \][/tex]
[tex]\[ y = -\frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{2}{3} \][/tex]

Since the coordinates are [tex]\((-2, -3)\)[/tex] and calculated y value is [tex]\( \frac{2}{3} \)[/tex], this point does not lie on the line (since [tex]\(-3 \neq \frac{2}{3}\)[/tex]).

Check (2, 5):
[tex]\[ y = \frac{2}{3}(2) + 2 \][/tex]
[tex]\[ y = \frac{4}{3} + 2 \][/tex]
[tex]\[ y = \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{10}{3} \][/tex]

Since the coordinates are [tex]\((2, 5)\)[/tex] and calculated y value is [tex]\( \frac{10}{3} \)[/tex], this point does not lie on the line (as [tex]\( 5 \neq \frac{10}{3}\)[/tex]).

Check (3, 4):
[tex]\[ y = \frac{2}{3}(3) + 2 \][/tex]
[tex]\[ y = 2 + 2 \][/tex]
[tex]\[ y = 4 \][/tex]

Since the coordinates are [tex]\((3,4)\)[/tex], this point lies on the line.

Check (6, 6):
[tex]\[ y = \frac{2}{3}(6) + 2 \][/tex]
[tex]\[ y = 4 + 2 \][/tex]
[tex]\[ y = 6 \][/tex]

Since the coordinates are [tex]\((6,6)\)[/tex], this point lies on the line.

Thus, the points Vera can use to graph the line are:

[tex]\[ (-3, 0), (3, 4), (6, 6) \][/tex]