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
y = 3 - s(x-4)
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.