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Sagot :
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]
[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]
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