how is this procedure derived?

A

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
962
Problem:
12310

I know this is simple and here is what I did:
12312

And here is what the solutions manual put forth:
12313)

What I want to know is how did the author derive that (x/5)+(y/-2)=1. I mean....one what? I am asking where this expression came from? Is it something that can be used with any two x, y intercepts? I tried it with another set of intercepts from a different graph and it didn't come out....maybe I screwed up some way, however (not unheard of). I have been looking for an explanation on Youtube and google and don't come up with anything like this. I know his solution and mine agree but I don't know how he got to it.
 

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That's called the two-intercept form for a line.

If we know a line has \(x\)-intercept at \((a,0)\) and \(y\)-intercept at \((0,b)\), then the slope is:

[MATH]m=-\frac{b}{a}[/MATH]
And so, using the slope-intercept form for a line, we may write the equation of the line through those two points as:

[MATH]y=-\frac{b}{a}x+b[/MATH]
Or:

[MATH]\frac{b}{a}x+y=b[/MATH]
Divide through by \(b\):

[MATH]\frac{x}{a}+\frac{y}{b}=1[/MATH]
And this is the formula we seek. :)
 
allegansveritatem said:
Problem:
View attachment 12310

I know this is simple and here is what I did:
View attachment 12312

And here is what the solutions manual put forth:
View attachment 12313)

What I want to know is how did the author derive that (x/5)+(y/-2)=1. I mean....one what? I am asking where this expression came from? Is it something that can be used with any two x, y intercepts? I tried it with another set of intercepts from a different graph and it didn't come out....maybe I screwed up some way, however (not unheard of). I have been looking for an explanation on Youtube and google and don't come up with anything like this. I know his solution and mine agree but I don't know how he got to it.
Click to expand...
That is an alternate representation of a line passing through (0,y1) and (x1,0) where x1 and y1 are x-intercept and y-intercept respectively. The equation for the line becomes:

(x/x1) + (y/y1) = 1
 
MarkFL said:
That's called the two-intercept form for a line.

If we know a line has \(x\)-intercept at \((a,0)\) and \(y\)-intercept at \((0,b)\), then the slope is:

[MATH]m=-\frac{b}{a}[/MATH]
And so, using the slope-intercept form for a line, we may write the equation of the line through those two points as:

[MATH]y=-\frac{b}{a}x+b[/MATH]
Or:

[MATH]\frac{b}{a}x+y=b[/MATH]
Divide through by \(b\):

[MATH]\frac{x}{a}+\frac{y}{b}=1[/MATH]
And this is the formula we seek. :)
Click to expand...
Thanks for the clear explanation. I have never come across that little trick before--but there is always something new coming up in this huge subject. I will copy your post and study it over again tomorrow when I am about to hit my text.
 
The last line of your written work should be 2x - 5y =10 not 2 (after multiplying both sides by 5 and rearranging).
 
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