Question

On a coordinate plane, a dashed straight line has a positive slope and goes through (negative 3, negative 3) and (3, 1). Everything below and to the right of the line is shaded.
Which linear inequality is represented by the graph?

y > Two-thirdsx – 2
y < Two-thirdsx + 2
y > Two-thirdsx + 1
y < Two-thirdsx – 1

Answer 1

Answer:

y < Two-thirdsx – 1

Step-by-step explanation:

use the information about the straight line to find equation of the line.

Given coordinates (-3,-3) and (3,1) find the slope

Slope, m=change in y/change in x

Change in y= 1--3=4

Change in x=3--3=6

m=4/6=2/3

Equation of line using m=2/3  and points (3,1) and (x,y) inform of y=mx+c

$\frac{y-1}{x-3} =\frac{2}{3} \\\\\\3(y-1)=2(x-3)\\\\\\3y-3=2x-6\\\\\\3y=2x-6+3\\\\\\3y=2x-3\\\\\\y=\frac{2}{3} x-1$

To write the inequality take a point in the shaded region and test it in the above equation

For example taking point (4,1)

$y=\frac{2}{3} x-1\\\\1=\frac{2}{3} *4-1\\\\\\1=\frac{8}{3} -1\\\\\\1=2.7-1\\\\\\1=1.7\\\\\\1<1.7\\\\\\y<\frac{2}{3} x-1$

Answer 2

The inequality is represented by  $y<\dfrac{2}{3}x-1$

Use the information about the straight line and find the equation of the line.

Given coordinates$(-3,-3)$ and $(3,1)$ find the slope

Slope,  $m=\dfrac{Change\;in\;y}{Change\;in\;x}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Change in  $y=1-(-3)=1+3=4$

Change in  $x=3-(-3)=3+3=6$

$\therefore m=\dfrac{4}{6}=\dfrac{2}{3}$

Equation of line using  $m=\frac{2}{3}$  and points $(3,1)$  and  $(x,y)$  in form of  $y=mx+c$

$\dfrac{y-1}{x-3}=\dfrac{2}{3}\\ \\ 3(y-1)=2(x-3)\\\\ 3y-3=2x-6\\ \\ 3y=2x-3\\ \\ y=\dfrac{2}{3}x -1$

To find the inequality take a point in the shaded region and check it in the above equation.

For example take point  $(2,-2)$

$y=\dfrac{2}{3}x-1\\ \\ -2=\dfrac{2}{3}*2-1\\ \\ -2=\dfrac{2*2-3}{3}\\ \\ -6=1\\ \\ -6< 1\\ \\ \therefore y< \dfrac{2}{3}x-1$

Learn more about Inequalities.

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