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Intercept Form: 3 Problems

jamescv31

New member
Nov 30, 2013
17
Greetings everyone, I need a help since the lecture of intercept form formula is \(\displaystyle X/A + Y/B = 1\) are provied as limited examples and couldn't find exact information regarding for that.

6) Reduce the equation \(\displaystyle 2x+3y = -1\) to intercept form.

7) find the equation of the line whose slope is 3x the y-intercept through (3,20)

8) Find the equation of the line through (6, -3) and whose intercepts add to 10.

Note: I'm trying to answer these questions however couldn't guarantee if its a correct.

Thank you.

(The intercept form that I'm referring is involved on the straight line)
 
Last edited:

soroban

Well-known member
Feb 2, 2012
409
Re: Intercept Form 3 Question Problems

Hello, jamescv31!

Intercept form: .[tex]\frac{x}{a} + \frac{y}{b} \:=\:1[/tex]

(6) Reduce the equation [tex]2x+3y \:=\: -1[/tex] to intercept form.

Multiply by -1: .[tex]-2x - 3y \:=\:1[/tex]

Therefore: .[tex]\frac{x}{\text{-}\frac{1}{2}} + \frac{y}{\text{-}\frac{1}{3}} \:=\:1[/tex]



(7) Find the equation of the line with slope 3 times the y-intercept
and through the point (3, 20).
[tex]\frac{x}{a} + \frac{y}{b} \:=\:1 \quad \Longleftrightarrow\quad bx + ay \:=\:ab[/tex] .[1]

The slope is: .[tex]m \,=\,-\frac{b}{a}[/tex]

We have: .[tex]-\frac{b}{a} \:=\:3b \quad\Rightarrow\quad a \,=\,-\tfrac{1}{3}[/tex]

Substitute into [1]: .[tex]bx - \tfrac{1}{3}y \:=\:\text{-}\tfrac{1}{3}b[/tex]

Substitute (3, 20): .[tex]3b - \tfrac{20}{3} \:=\:\text{-}\tfrac{1}{3}b \quad\Rightarrow\quad b = 2[/tex]

Therefore: .[tex]\frac{x}{\text{-}\frac{1}{3}} + \frac{y}{2} \:=\:1[/tex]




(8) Find the equation of the line through (6, -3)
and whose intercepts add to 10.
There are two solutions.


We have: .[tex]bx + ay \:=\:ab[/tex]

Substitute (6,-3): .[tex]6b - 3a \:=\:ab[/tex] .[1]

We have: .[tex]a + b \:=\:10 \quad \Rightarrow\quad b \:=\:10-a[/tex]

Substitute into [1]: .[tex]6(10-a) - 3a \:=\:a(10-a) [/tex]

. . [tex]60 - 6a - 3a \:=\:10a - a^2[/tex]

. . [tex]a^2 - 19a + 60\:=\:0 \quad \Rightarrow\quad (a-4)(a-15) \:=\:0[/tex]

Hence: .[tex]\begin{Bmatrix}a&=&4 \\ b&=&6\end{Bmatrix}\quad\begin{Bmatrix}a&=&15 \\ b &=&\text{-}5 \end{Bmatrix}[/tex]

Therefore: .[tex]\frac{x}{4} + \frac{y}{6} \:=\:1\;\text{ and }\; \frac{x}{15} - \frac{y}{5} \:=\:1[/tex]
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Intercept Form 3 Question Problems

Hello and welcome to MHB, jamescv31. (Sun)

I need to mention to you that we ask that no more than two questions be asked in a single thread. This helps prevent a thread from potentially becoming convoluted and hard to follow in the case where more than one helper may be trying to help with different problems at the same time. This helps keep MHB more organized and useful for everyone. (Nerd)

Best Regards,

Mark.
 

jamescv31

New member
Nov 30, 2013
17
Re: Intercept Form 3 Question Problems

MarkFL: Sorry about that, well next time I'm gonna post a question which is really hard for me to understand very well prior for the rules n this forum. :)
 

jamescv31

New member
Nov 30, 2013
17
Re: Intercept Form 3 Question Problems

On number 7) the given of the slope is 3x, my mistaken to posted it but is there a difference on the answer?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Intercept Form 3 Question Problems

On number 7) the given of the slope is 3x, my mistaken to posted it but is there a difference on the answer?
The slope of a straight line cannot vary, it must be constant. :D
 

jamescv31

New member
Nov 30, 2013
17
Re: Intercept Form 3 Question Problems

Yes, so it means "3x" or "3 times" are the same though on the given y-intercept?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Intercept Form 3 Question Problems

Yes, so it means 3x or 3 times are the same though on the given y-intercept?
I believe soroban correctly interpreted the problem to say the value of the line's slope is 3 times that of the line's $y$-intercept.
 

jamescv31

New member
Nov 30, 2013
17
Re: Intercept Form 3 Question Problems

\(\displaystyle \displaystyle 3b - \tfrac{20}{3} \:=\:\text{-}\tfrac{1}{3}b \quad\Rightarrow\quad b = 2\)

To get the answer of \(\displaystyle \displaystyle \frac{x}{\text{-}\frac{1}{3}} + \frac{y}{2} \:=\:1\)

I have a question regarding to this solution on number 7: where did the b = 2 came from? I've already understand the rest.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Intercept Form 3 Question Problems

\(\displaystyle \displaystyle 3b - \tfrac{20}{3} \:=\:\text{-}\tfrac{1}{3}b \quad\Rightarrow\quad b = 2\)

To get the answer of \(\displaystyle \displaystyle \frac{x}{\text{-}\frac{1}{3}} + \frac{y}{2} \:=\:1\)

I have a question regarding to this solution on number 7: where did the b = 2 came from? I've already understand the rest.
We have:

\(\displaystyle 3b-\frac{20}{3}=-\frac{1}{3}b\)

Multiply through by 3 to obtain:

\(\displaystyle 9b-20=-b\)

Add $b+20$ to both sides:

\(\displaystyle 10b=20\)

Divide through by 10:

\(\displaystyle b=2\)