Need help finding the slope of a line for a triangle area problem?

[mindauggas]

Homework Statement

Find the possible slopes of a line that passes through (4,3) so that the portion of the line in the first quadrant forms a triangle of area 27 with the positive coordinate axes.

Homework Equations

Not given, but I think S(triangle)=ah/2 and point-slope form of the equation of a line will come in handy.

The Attempt at a Solution

I notice that since $S^{(triangle)}=\frac{ah}{2}$ and S=27, ah=54. Because h is the y coordinate and a is the length of the triangle (but not the x coordinate - since the triangle might not pass trough the origin) y=h.

To find x coordinate I mark the x of the point (x,0) (where the line intersects the x axis) - x$_{1}$, and the x at y=h as x$_{2}$. Now x$_{2}$=x$_{1}$+a.

I noticed that y$_{(at:y=h)}$*a=54

I can also us the point (x$_{1}$,0) and the given point (4,3) to find an expression of the slope $m=\frac{3}{4-x}$.

Stuck here, I do not know how to glue this information together or is it really necessary ... please help.

 

[Sourabh N]

First step would be to draw a triangle that (roughly) satisfies the conditions given in the question - passing through (4,3) and positive intercepts on x,y axes.

The general equation of a line on your plane involves two variables, slope and intercept. Since you know one point that lies on your line, you can eliminate one of the variables. Now you can find the y-intercept and the x-intercept and consequently the area, as a function of the remaining variable.

 

[mindauggas]

First step would be to draw a triangle that (roughly) satisfies the conditions given in the question - passing through (4,3) and positive intercepts on x,y axes.

The general equation of a line on your plane involves two variables, slope and intercept. Since you know one point that lies on your line, you can eliminate one of the variables.

If the indicated equation is y=mx+b and the point (4,3) i can write 3=m4+b. But what do you mean by eliminating one of the variables?

 

[Sourabh N]

If the indicated equation is y=mx+b and the point (4,3) i can write 3=m4+b. But what do you mean by eliminating one of the variables?

After some rearranging, the above equation gives 'b' in terms of 'm'. Thus you can replace every instance of 'b' by this function. Now everything is a function of 'm'.

^You eliminated one variable!

 

[mindauggas]

Ok, so I have b=-m4+3. But this is the y intercept and I actually can't see how it can help. When I tried to solve the problem I tried to use the point (x,0) (so the x intercept). My logic was, that the y intecept in this case is below x-axis so it does not comprise any of the "relevant points" of the tringle.

So the question now is how to use this equation to "find the y-intercept and the x-intercept and consequently the area, as a function of the remaining variable" ?

 

[Sourabh N]

Ok, so I have b=-m4+3. But this is the y intercept and I actually can't see how it can help. When I tried to solve the problem I tried to use the point (x,0) (so the x intercept). My logic was, that the y intecept in this case is below x-axis so it does not comprise any of the "relevant points" of the tringle.

Can you explain how you arrive at the conclusion "the y intecept in this case is below x axis"?

 

[mindauggas]

Can you explain how you arrive at the conclusion "the y intecept in this case is below x axis"?

By "in this case" i meant "in this exercise".

I didn't arrive at it ... I postulate it, assume it. There is only one argument that guides my assumption - if when x=0 y is positive then the triangle that results will extend to the 2nd quadrant, contrary to the given information. So I should correct myself, I assumed that when x=0 y is non-negative (it can also be zero - so not necessarily below x axis).

 

[Sourabh N]

By "in this case" i meant "in this exercise".

I didn't arrive at it ... I postulate it, assume it. There is only one argument that guides my assumption - if when x=0 y is positive then the triangle that results will extend to the 2nd quadrant, contrary to the given information.

How is this contrary to given information? (The question in post #1 says the line must have some portion in first quadrant. No conditions pertaining to other quadrants is given.)

So I should correct myself, I assumed that when x=0 y is non-negative (it can also be zero - so not necessarily below x axis).

That seems almost right. y MUST be positive at x = 0, otherwise (quoting from post #1) "the portion of the line in the first quadrant forms a triangle of area 27 with the positive coordinate axes." will not be satisfied.

 

[mindauggas]

How is this contrary to given information? (The question in post #1 says the line must have some portion in first quadrant. No conditions pertaining to other quadrants is given.)

I presumed that not a portion, but the whole triangle has to be in the first quadrant ... because if the triangle extended to the 2nd quadrant the shape in the first quadrant would not form a triangle, and that contradicts (as I understand it, at least - maybe I'm wrong) "that the portion of the line in the first quadrant forms a triangle" (first post).

 

[Sourabh N]

The triangle will be formed by the line, the positive x-axis and the positive y-axis. I said (actually I rephrased post #1) the line must have some portion in the first quadrant so that it is possible to make a triangle. If it confuses you, you can safely ignore that part of my reply

I suggest you draw a few lines that satisfy the conditions given in the problem to get an idea.

 

[mindauggas]

The triangle will be formed by the line, the positive x-axis and the positive y-axis. I said (actually I rephrased post #1) the line must have some portion in the first quadrant so that it is possible to make a triangle. If it confuses you, you can safely ignore that part of my reply

I suggest you draw a few lines that satisfy the conditions given in the problem to get an idea.

How can such graph form a triangle in the first quadrant?

What should I do next?

 

[Sourabh N]

How can such graph form a triangle in the first quadrant?

What should I do next?

That does not satisfy all the conditions given in the question, does it?

1. The general equation of a line passing is y = mx + b.
2. Since the line passes through (4,3), we have 3 = 4m + b which implies b = 3 - 4m.
3. The equation of line now is y = mx + 3 - 4m.

^This is what I meant when I said "You eliminated one variable!"

 

[SammyS]

How can such graph form a triangle in the first quadrant?
What should I do next?

The graph you show forms a triangle in the second quadrant, when combined with the coordinate axes.

What must be true of the x and y intercepts if the line and coordinate axes are to form a triangle in the first quadrant ?

How are the intercepts related to the number, 27 ?

 

[mindauggas]

That does not satisfy all the conditions given in the question, does it?

Well of course it does not ... therefore my corrected "presumption"/"assumption" was correct.

The graph you show forms a triangle in the second quadrant, when combined with the coordinate axes.

What must be true of the x and y intercepts if the line and coordinate axes are to form a triangle in the first quadrant ?

As was said previously both x and y intercepts must have non-negative values.

How are the intercepts related to the number, 27 ?

I don't know how they are related to the intercepts ... I know that the point (x,0) and (x+a,y) comprise a line that satisfies the given conditions and since S(traingle)=ah/2 (a ia a side and h - hypotenuse). I can deduce that ah/2=27 therefore ah=54. Since y in the coordinate (x+a,y) is definitely equal to h therefore ay=54.

But, I don't know how they are related to the intercepts...

Might the equation derived with the help of Sourabh N help? b = 3 - 4m (the y intercept of this line expressed as a function of the slope).

 

[SammyS]

How can such graph form a triangle in the first quadrant?

What should I do next?

Obviously such a graph doesn't form a triangle in the first quadrant.

As the problem implies, the x and y intercepts must be positive. Therefore, the slope of the line is negative.

The line and the axes form a right triangle which makes calculation of the area of the triangle straight forward using the values from the x and y intercepts.

 

[mindauggas]

Sorry, I completely missed the point of the problem:

So, now I know, that x and y intercept values multiplied amount to 54 (27*2), the slope is negative (2), passes trough a point (4,3) (3), the slope can be written as $m=\frac{3-0}{4-x}$ (between the points (4,3) and (x,0) (the x intercept) (or it can be written on a point (0,y) fot the y intercept.

Lost at this point ...

 

[SammyS]

Sorry, I completely missed the point of the problem:

So, now I know, that x and y intercept values multiplied amount to 54 (27*2), the slope is negative (2), passes trough a point (4,3) (3), the slope can be written as $m=\frac{3-0}{4-x}$ (between the points (4,3) and (x,0) (the x intercept) (or it can be written on a point (0,y) fot the y intercept.

Lost at this point ...

Now, look at Post # 12 again to see what Sourabh N suggested.

 

[mindauggas]

Now, look at Post # 12 again to see what Sourabh N suggested.

His equation was: y=mx+3-4m

This becomes:

$y=\frac{3}{4-x}(x)+3-4\frac{3}{4-x}$ (?)

Then $y=\frac{3}{4-x}(x)-1$

This is the slope-intercept form, so b=-1, but this is clearly absurd ... any suggestions?

 

[Sourabh N]

His equation was: y=mx+3-4m

This becomes:

$y=\frac{3}{4-x}(x)+3-4\frac{3}{4-x}$ (?)

Then $y=\frac{3}{4-x}(x)-1$

This is the slope-intercept form, so b=-1, but this is clearly absurd ... any suggestions?

^You made a mistake in going from the first equation to the second equation there.

Remember what the question is: Find the possible slopes. The equation of line in #12 is in terms of the slope (represented by m). Your next step should be to obtain x and y intercepts in terms of this slope m.

 

[mindauggas]

^You made a mistake in going from the first equation to the second equation there.

Remember what the question is: Find the possible slopes. The equation of line in #12 is in terms of the slope (represented by m). Your next step should be to obtain x and y intercepts in terms of this slope m.

$x_{intercept}$ 0=mx+3-4m >> $-3=m(x-4)$ >> $\frac{-3}{x-4}=m$ (i don't see any way where the variable x would be on the left by itself ... is there one?)

$y_{intercept}$ y=3-4m

Is this correct? Whats the next step?

 

[Sourabh N]

Almost there!

$x_{intercept}$ 0=mx+3-4m >> $-3=m(x-4)$ >> $\frac{-3}{x-4}=m$ (i don't see any way where the variable x would be on the left by itself ... is there one?)

Instead of moving 3 over to the left side, move mx. Then divide both sides by m and you'll find x on the left side.

$y_{intercept}$ y=3-4m

Is this correct? Whats the next step?

This looks perfect.
You figured out the next step in #16, as quoted below:

So, now I know, that x and y intercept values multiplied amount to 54 (27*2),

 

[mindauggas]

$y=3-4m$

$0=mx+3-4m$
$-mx=3-4m$ (dividing by -m)
$x=\frac{-3}{m}+4$

$xy=54$, then

$(\frac{-3}{m}+4)(3-4m)=54$
$\frac{-9}{m}+4m*\frac{-3}{m}+12-16m=54$
$\frac{-9}{m}-16m=30$ (multiplying by m, we get a quadratic)
$16m^{2}+30+9=0$
With the roots -3/2 and -3/8

Awesomenessness ... Thank you very much your patience :)

 

[Sourabh N]

Right. Now multiply both sides by m. You will arrive at an equation quadratic in m. Solving this will give you the possible values of slope.
As a verification, you should check if the sign of the solutions is negative, as SammyS suggested.

 

[mindauggas]

Yes ... Solved ... Thank's again :)

 

[Sourabh N]

No problem