Question

Two collinear points on a line are given in the table below:

Answer 1

Answer:

$(4,3)$ and $(7,2)$ do not lie on the line

Step-by-step explanation:

Given

$(0,0)\ and\ (2,1)$

Required

Determine which points that are not on the line

First, we need to determine the slope (m) of the line:

$m = \frac{y_2 - y_1}{x_2- x_1}$

Where

$(x_1,y_1) = (0,0)$

$(x_2,y_2) = (2,1)$

So;

$m = \frac{y_2 - y_1}{x_2- x_1}$

$m = \frac{1 - 0}{2-0}$

$m = \frac{1}{2}$

Next, we determine the line equation using:

$y - y_1 = m(x -x_1)$

Where

$m = \frac{1}{2}$

$(x_1,y_1) = (0,0)$

$y - y_1 = m(x -x_1)$ becomes

$y - 0 = \frac{1}{2}(x - 0)$

$y = \frac{1}{2}x$

To determine which point is on the line, we simply plug in the  values of x to in the equation check.

For $(4,2)$

$x = 4$ and $y =2$

Substitute 4 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 4$

$2 = \frac{4}{2}$

$2=2$

This point is on the graph

For $(4,3)$

$x = 4$ and $y = 3$

Substitute 4 for x and 3 for y in $y = \frac{1}{2}x$

$3 = \frac{1}{2} * 4$

$3 = \frac{4}{2}$

$3 \neq 2$

This point is not on the graph

For $(7,2)$

$x = 7$ and $y = 2$

Substitute 7 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 7$

$2 = \frac{7}{2}$

$2 \neq 3.5$

This point is not on the graph

$(\frac{4}{8},\frac{2}{8})$

$x = \frac{4}{8}$ and $y = \frac{2}{8}$

Substitute $\frac{4}{8}$ for x and $\frac{2}{8}$ for y in $y = \frac{1}{2}x$

$\frac{2}{8} = \frac{1}{2} * \frac{4}{8}$

$\frac{2}{8} = \frac{1 * 4}{8 * 2}$

$\frac{2}{8} = \frac{4}{16}$

$\frac{1}{4} = \frac{1}{4}$

This point is on the graph