all points (x,y) that lie below the line

[urooj.khan]

[stock.mojo11]

The line passes through (6,0) and (0,3)

Slope= 3/-6 = -1/2

y = (-1/2) x + b

y intercept b is the value at which x=0

3= 0 +b

eq of line is y = ( -1/2 ) x + 3

Q is asking for an inequality for all the points beneath (read beneath less than)

y < ( -1/2 ) x + 3

[RonPurewal]

dirty problem.

the diagram strongly suggests that we should only consider points located in the first quadrant.
if we only consider points in the first quadrant, then those points satisfy all three of inequalities (a), (d), (e).
to eliminate the former two, you have to consider points that lie beneath the given line but outside the first quadrant.

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the best way to do this problem is to write the equation of the line in slope-intercept form, as has been done by the poster above me.

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if you don't know how to do this, or you've forgotten how, then you can also plug in points.

the key is that you should PLUG IN POINTS FROM DIFFERENT AREAS OF THE GIVEN REGION.

try the point (6, -1), which is guaranteed to lie within the given region [since (6, 0) is on the line].
this point doesn't satisfy (b) or (c), so eliminate those.

try the point (-1, 3), which is guaranteed to lie within the given region [since (0, 3) is on the line].
this point doesn't satisfy (a) or (d), so eliminate those.

(e) is the last man standing.

[sachin.w]

eq of line is y = ( -1/2 ) x + 3

Q is asking for an inequality for all the points beneath (read beneath less than)

y < ( -1/2 ) x + 3

I don't understand how we can conclude that for the points below the line above mentioned inequality is the correct answer.

Only way to ascertain I believe would be to plug in numbers. Or is there any concept I am missing ?

[RonPurewal]

I don't understand how we can conclude that for the points below the line above mentioned inequality is the correct answer.

Only way to ascertain I believe would be to plug in numbers. Or is there any concept I am missing ?

well, "y" is the vertical coordinate. so, if you see y < blah, then that's always going to be the stuff that lies under the graph of "blah". this is what it means for a vertical coordinate to be less than something -- it's lower down.

of course, testing points is not particularly painful or time-consuming. so, if you are not immediately clear on this concept, you can easily test some points.

[sachin.w]

I don't understand how we can conclude that for the points below the line above mentioned inequality is the correct answer.

Only way to ascertain I believe would be to plug in numbers. Or is there any concept I am missing ?

well, "y" is the vertical coordinate. so, if you see y < blah, then that's always going to be the stuff that lies under the graph of "blah". this is what it means for a vertical coordinate to be less than something -- it's lower down.

Hi Ron Sir,
I am still not convinced. Could you please let us know how this is true or how this can be derived?
This is one concept I never learnt in my school. So finding it difficult to simply accept it .

[jnelson0612]

I don't understand how we can conclude that for the points below the line above mentioned inequality is the correct answer.

Only way to ascertain I believe would be to plug in numbers. Or is there any concept I am missing ?

well, "y" is the vertical coordinate. so, if you see y < blah, then that's always going to be the stuff that lies under the graph of "blah". this is what it means for a vertical coordinate to be less than something -- it's lower down.

Hi Ron Sir,
I am still not convinced. Could you please let us know how this is true or how this can be derived?
This is one concept I never learnt in my school. So finding it difficult to simply accept it .

Let's graph a line so you can see what Ron is talking about.

Take the equation y < x + 2. The fundamental equation for a line is y=mx + b. b is the y-intercept of the line, the number in front of x gives you the slope of the line, and then x and y are the individual points.

Let's start with x=0. If we plug x=0 into our equation, then y < 2. y may be any value *below* 2 on the y access.

If x=1, then y < 3. y may be any value below 3 at the point of x=1. Now that you have two points go ahead and plot them on the graph, draw a line, and extend the line out both ways. Look at what y is allowed to be. It is allowed to be everything *below* that line. Just plug numbers in if you have any doubt and please let us know if you have further questions.