Answer:
[tex](4,3)[/tex] and [tex](7,2)[/tex] do not lie on the line
Step-by-step explanation:
Given
[tex](0,0)\ and\ (2,1)[/tex]
Required
Determine which points that are not on the line
First, we need to determine the slope (m) of the line:
[tex]m = \frac{y_2 - y_1}{x_2- x_1}[/tex]
Where
[tex](x_1,y_1) = (0,0)[/tex]
[tex](x_2,y_2) = (2,1)[/tex]
So;
[tex]m = \frac{y_2 - y_1}{x_2- x_1}[/tex]
[tex]m = \frac{1 - 0}{2-0}[/tex]
[tex]m = \frac{1}{2}[/tex]
Next, we determine the line equation using:
[tex]y - y_1 = m(x -x_1)[/tex]
Where
[tex]m = \frac{1}{2}[/tex]
[tex](x_1,y_1) = (0,0)[/tex]
[tex]y - y_1 = m(x -x_1)[/tex] becomes
[tex]y - 0 = \frac{1}{2}(x - 0)[/tex]
[tex]y = \frac{1}{2}x[/tex]
To determine which point is on the line, we simply plug in the values of x to in the equation check.
For [tex](4,2)[/tex]
[tex]x = 4[/tex] and [tex]y =2[/tex]
Substitute 4 for x and 2 for y in [tex]y = \frac{1}{2}x[/tex]
[tex]2 = \frac{1}{2} * 4[/tex]
[tex]2 = \frac{4}{2}[/tex]
[tex]2=2[/tex]
This point is on the graph
For [tex](4,3)[/tex]
[tex]x = 4[/tex] and [tex]y = 3[/tex]
Substitute 4 for x and 3 for y in [tex]y = \frac{1}{2}x[/tex]
[tex]3 = \frac{1}{2} * 4[/tex]
[tex]3 = \frac{4}{2}[/tex]
[tex]3 \neq 2[/tex]
This point is not on the graph
For [tex](7,2)[/tex]
[tex]x = 7[/tex] and [tex]y = 2[/tex]
Substitute 7 for x and 2 for y in [tex]y = \frac{1}{2}x[/tex]
[tex]2 = \frac{1}{2} * 7[/tex]
[tex]2 = \frac{7}{2}[/tex]
[tex]2 \neq 3.5[/tex]
This point is not on the graph
[tex](\frac{4}{8},\frac{2}{8})[/tex]
[tex]x = \frac{4}{8}[/tex] and [tex]y = \frac{2}{8}[/tex]
Substitute [tex]\frac{4}{8}[/tex] for x and [tex]\frac{2}{8}[/tex] for y in [tex]y = \frac{1}{2}x[/tex]
[tex]\frac{2}{8} = \frac{1}{2} * \frac{4}{8}[/tex]
[tex]\frac{2}{8} = \frac{1 * 4}{8 * 2}[/tex]
[tex]\frac{2}{8} = \frac{4}{16}[/tex]
[tex]\frac{1}{4} = \frac{1}{4}[/tex]
This point is on the graph
