The equation of the line passing through the point $(3,15)$ and the origin can be found using the slope-intercept form $y = mx+b$, where $m$ is the slope of the line.
The slope of the line can be found using the formula $m = \frac{y_2-y_1}{x_2-x_1}$, where $(x_1,y_1) = (0,0)$ and $(x_2,y_2) = (3,15)$.
$m = \frac{15-0}{3-0} = \frac{15}{3} = 5$.
Therefore, the equation of the line is $y = 5x$.
To determine which points could be on the graph of the line, substitute the $x$-coordinate of each point into the equation $y=5x$ and see if the resulting $y$-coordinate matches the $y$-coordinate of the given point.
For point A: $x=1$, $y=5(1)=5$. The $y$-coordinate doesn't match, so point A is not on the graph.
For point B: $x=2$, $y=5(2)=10$. The $y$-coordinate matches, so point B is on the graph.
For point C: $x=2$, $y=5(2)=10$. The $y$-coordinate matches, so point C is on the graph.
For point D: $x=6$, $y=5(6)=30$. The $y$-coordinate doesn't match, so point D is not on the graph.
For point E: $x=7$, $y=5(7)=35$. The $y$-coordinate doesn't match, so point E is not on the graph.
For point F: $x=9$, $y=5(9)=45$. The $y$-coordinate matches, so point F is on the graph.
For point G: $x=8$, $y=5(8)=40$. The $y$-coordinate doesn't match, so point G is not on the graph.
Therefore, the points that could also be on the graph of the line are B, C, and F.
A student graphs a line that passes through the point $\left(3,\ 15\right)$ and the origin.
Which points could also be on the graph of the line?
Select all that apply.
Responses
A $\left(1,\ 3\right)$$\left(1,\ 3\right)$
B $\left(2,\ 10\right)$$\left(2,\ 10\right)$
C $\left(2,\ 14\right)$$\left(2,\ 14\right)$
D $\left(6,\ 30\right)$$\left(6,\ 30\right)$
E $\left(7,\ 21\right)$$\left(7,\ 21\right)$
F $\left(9,\ 45\right)$$\left(9,\ 45\right)$
G $\left(8,\ 20\right)$$\left(8,\ 20\right)$
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