Question

The table shown lists random points found on a coordinate plane.

Answer 1

The two lines created from the points (0, 10), (2, 7) and (-3, -3) works

exclusively with the three points as presented in the the attached graph.

Which methods can be used to obtain linear functions?

The possible value in the table obtained from a similar question is presented as follows;

$\begin{array}{|lcl|}(-1, \, 1)&&(2, \, 7)\\(-4, \, 8)&&(2, \, 11)\\(0, \, 10)&&(-3, \, -3)\end{array}\right]$

A line is defined as the shortest distance between two points.

Taking the points (0, 10) and (-3, -3), we have;

$Slope = \dfrac{10 - (-3)}{0 - (-3)} = \dfrac{13}{3}$

Which gives;

$y -10 = \dfrac{13}{3} \times x$

$y = \dfrac{13}{3} \cdot x + 10$

From the attached graph, the points (-1, 1), (2, 7), (2, 11), and (-4, 8) are not on the line.

The linear function, $y = \dfrac{13}{3} \cdot x + 10$ works exclusively for the points (0, 10) and (-3, -3), given that the coordinates of points on the line are; (-1, $5\frac{2}{3}$), (2, $18\frac{2}{3}$), (-4, $-7\frac{1}{3}$)

Second line

Taking the points (2, 11) and (-3, -3), we have;

$Slope = \mathbf{\dfrac{11 - (-3)}{2 - (-3)}}= \dfrac{14}{5} = 2.8$

Which gives;

$y - (-3) = 2.8\times (x - (-3)) = 2.8 \cdot x + 8.4$

y = 2.8·x + 8.4 - 3 = 2.8·x + 5.4

Which gives;

y = 2.8·x + 5.4

The line, y = 2.8·x + 5.4, works exclusively for the points (2, 11) and (-3, -3), given that the points (2, 7), (2, 11), (0, 10), and (-3, -3) are not on the line

The coordinates of points on the line are; (-1, 2.6), (-4, -5.8), and (0, 5.4).

Learn more about linear functions here:

brainly.com/question/20478559