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

c)
`y\leq (2)/(3)x+(1)/(5)`

Explanation:

Given points:

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

Finding slope:

`m=(y_2-y_1)/(x_2-x_1)=(2.2-0.2)/(3-0)=(2)/(3)`

Using point intercept equation to find equation of line:

`(y-y_1)=m(x-x_1)\\(y-0.2)=(2)/(3)(x-0)\\`

`(y-0.2)=(2)/(3)x`

Simplifying to standard form by adding 0.2 to both sides.

`y-0.2+0.2=(2)/(3)x+0.2`

`y=(2)/(3)x+0.2`

`y=(2)/(3)x+(1)/(5)` [∵
`0.2=(2)/(10)=(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 (2)/(3)x+(1)/(5)` [∵
`\leq` represent shaded are below the line which lies to the right including the line(solid)]

Answer 2

`y\leq (2x)/(3)+(1)/(5)`

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)=(y1-y2)/(x1-x2)*(x-x2)`

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

`y-0.2=(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 (2x)/(3)+(1)/(5)`