Slope of Linear function

[Ibraheem]

Hello,
I'm having a hard time solving a linear function slope problem.So I would be thankful if someone could provide me with an answer and explanation of the problem.

The problem is the following: 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.

 

[BvU]

To determine the triangle area you will need to calculate where the line crosses the axes.
Start with a drawing. Give the general form of a line through (4,3) with slope m and solve x for y = 0 and y for x = 0.

 

[Chestermiller]

Are you able to see how a line can form the hypotenuse of a right triangle with the positive coordinate axes?

Chet

 

[Ibraheem]

I've tried to find where the line crosses the axes, but I ended up with many variables with no answer.

I would really appreciate it if you could provide the answer with an explanation.

 

[Chestermiller]

Show us the details of what you did. Then we can help you. That's what this is all about.

Chet

 

[Ibraheem]

Sorry for not posting my solution attempt. I'm Kind of new to this website.

I assumed that one of the possible lines crosses the x-axis at (a,0) and the y-axis at (0,b) and that a*b=54 since (Area of triangle*2 )=27*2=ab. I used the the two-intercept form of the line equation.

The solution attempt: (-b*x/a)+b=y → bx+ay=ab →4b+3a=54
this is where I ended up
I tried to assigning numbers to satisfy the equation 4b+3a=54 , and the results I got was( a=12 and b=4.5) and (b=9 and a=6)
so the slopes are (-3/2) and (-3/8)

I don't know if there is any other possible slopes or how to make sure there isn't. Also, is there a way to find the slopes without ending up assigning numbers to satisfy 4b+3a=ab; that is, a way to find the answer without dealing with two variables.

 

[Chestermiller]

Sorry for not posting my solution attempt. I'm Kind of new to this website.

I assumed that one of the possible lines crosses the x-axis at (a,0) and the y-axis at (0,b) and that a*b=54 since (Area of triangle*2 )=27*2=ab. I used the the two-intercept form of the line equation.

The solution attempt: (-b*x/a)+b=y → bx+ay=ab →4b+3a=54

You were very close to having it solved.

You can still use ab= 54 again by substituting for either b or a in the above equation. This will give you an equation exclusively in terms of either a or b.

Chet

 

[Ray Vickson]

Sorry for not posting my solution attempt. I'm Kind of new to this website.

I assumed that one of the possible lines crosses the x-axis at (a,0) and the y-axis at (0,b) and that a*b=54 since (Area of triangle*2 )=27*2=ab. I used the the two-intercept form of the line equation.

The solution attempt: (-b*x/a)+b=y → bx+ay=ab →4b+3a=54
this is where I ended up
I tried to assigning numbers to satisfy the equation 4b+3a=54 , and the results I got was( a=12 and b=4.5) and (b=9 and a=6)
so the slopes are (-3/2) and (-3/8)

I don't know if there is any other possible slopes or how to make sure there isn't. Also, is there a way to find the slopes without ending up assigning numbers to satisfy 4b+3a=ab; that is, a way to find the answer without dealing with two variables.

Those are the only two solutions. The easiest way to see this is to use a single variable (slope = ##-s##), and write the equation of the line as
[tex] y = 3 - s(x-4)[/tex]
Note that when ##x = 4## we have ##y = 3##, as we need; and the slope is ##-s##, as stated. You can find the x- and y-intercepts in terms of ##s##: to find the x-intercept ##B## (= "base"), set ##y = 0## and solve for ##x## in terms of ##s##. To find the y-intercept ##H## (= "height"), just put ##x = 0##. Now the area ##A = \frac{1}{2} BH## becomes a function of ##s##. Setting ##A = 27## yields a quadratic equation for ##s##, so has at most two roots. In this case it does have exactly two positive, real roots that you can find using the quadratic-root formula.