Head scratcher: Find "a" so ax+4y=3, x/(a-1)+2y-1=0 are perpendicular

[bjarni86]

[FONT=&quot]I am scratching my way through my scalp over this... can somebody please help?

You have 2 equations to lines that are perpendicular to each other. [/FONT]

[FONT=&quot]ax+4y=3
[/FONT]
[FONT=&quot]and
[/FONT]
[FONT=&quot]x/(a-1)+2y-1=0[/FONT]

[FONT=&quot]Solve for a so these lines are perpendicular.[/FONT]

 

[mmm4444bot]

You have 2 equations to lines that are perpendicular to each other.

ax+4y=3

and

x/(a-1)+2y-1=0

Solve for a so these lines are perpendicular.

What have you tried or thought about, so far? That is, where did you get stuck?

My approach would begin by putting each equation into Slope-Intercept Form. Each of the slopes will be an expression containing a.

What have you learned about the relationship between slopes of perpendicular lines?

Use that fact, to write an equation which you can solve for a. :cool:

PS: Please be sure that you've read the forum guidelines. Thanks.

 

[bjarni86]

Thanks!

So, I've tried to isolate y in the first equation and I've come up with y=3-ax/4

As for the second one, I dont know what to do with the fraction x/a-1. The only simplification that comes to mind is to eliminate the -1.

Overall, I am not getting the bigger picture of how I am going to determine a or the slope from my findings in the first equation as it is not quite in the standard slope intersect form...

Thanks again!

 

[Deleted member 4993]

Thanks!

So, I've tried to isolate y in the first equation and I've come up with y=3-ax/4

As for the second one, I dont know what to do with the fraction x/a-1. The only simplification that comes to mind is to eliminate the -1.

Overall, I am not getting the bigger picture of how I am going to determine a or the slope from my findings in the first equation as it is not quite in the standard slope intersect form...

Thanks again!

Write the first equation in the form:

y = m₁ * x + b₁

Write the second equation in the form:

y = m₂ * x + b₂


For the lines to be perpendicular, you must have:

m₁ * m₂ = -1

 

[mmm4444bot]

So, I've tried to isolate y in the first equation and I've come up with y=3-ax/4

That's correct, but let's show it in the form y=m*x+b, with x factored out, so that we can see the expression for slope:

y = -a/4 * x + 3

The slope is m = -a/4

As for the second one, I dont know what to do with the fraction x/(a-1).

Treat that ratio as a single object, for now; that is, you can add or subtract x/(a-1) on each side of the equation, as needed. You can multiply it by another ratio, etc.

After you solve for y, factor out the x, to obtain an expression for the slope (like I did above, with -ax/4).

Once you have the two slope expressions, use the relationship for slopes of perpendicular lines, to write an equation. The slopes of perpendicular lines are negative reciprocals of each other (i.e., their product is -1, as Subhotosh showed). Solve that equation for a. :cool: