Question

On a coordinate plane, a solid straight line has a positive slope and goes through (0, 0.2) and (3, 2.2). Everything to the right of the line is shaded.
Which linear inequality is represented by the graph?

y > Two-thirdsx – One-fifth
y ≥ Three-halvesx + One-fifth
y ≤ Two-thirdsx + One-fifth
y < Three-halvesx – One-fifth

Answer 1

Answer:

c) $y\leq \frac{2}{3}x+\frac{1}{5}$

Step-by-step explanation:

Given points:

$(0,0.2)$ and $(3,2.2)$

Finding slope:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{2.2-0.2}{3-0}=\frac{2}{3}$

Using point intercept equation to find equation of line:

$(y-y_1)=m(x-x_1)\\(y-0.2)=\frac{2}{3}(x-0)$

$(y-0.2)=\frac{2}{3}x$

Simplifying to standard form by adding 0.2 to both sides.

$y-0.2+0.2=\frac{2}{3}x+0.2$

$y=\frac{2}{3}x+0.2$

$y=\frac{2}{3}x+\frac{1}{5}$   [∵ $0.2=\frac{2}{10}=\frac{1}{5}$ ]

Now, the graph has right side shaded with the solid line. Thus we can write the inequality represented by the graph as:

$y\leq \frac{2}{3}x+\frac{1}{5}$ [∵ $\leq$  represent shaded are below the line which lies to the right including the line(solid)]

Answer 2

Answer:

$y\leq \frac{2x}{3}+\frac{1}{5}$

Step-by-step explanation:

It is given that we have a line equation from (0,0.2) to (3, 2.2) with positive slope.Firstly for calculating the line equation between two points (x1,y1),(x2,y2) we have the relation ,

$(y-y2)=\frac{y1-y2}{x1-x2}*(x-x2)$

In the given case substituting them we get the line equation as

$y-0.2=\frac{2.2-0.2}{3-0}*(x-0 )\\2x-3y+0.6=0$

The condition was that everything to the right of the curve is shaded.

This is an inequality which needs to be solved with boundary conditions.

We notice that for x to the right of the equation y is always less than the existing line.(As it has a positive slope)

So for all x greater than or to the right of the line y lies below the line.

$y\leq \frac{2x}{3}+\frac{1}{5}$