Please help!

[pinkcalculator]

We've gotten a list of problems for my class, and I'm really struggling with this one.

Suppose the line ax-4y=9 is perpendicular to the line 8x=-11-3y. What is the value of a. At what exact point do these two lines intersect?

I know that you have to find the slopes (m) and that if the lines are perpendicular, the slopes will be negative reciprocals.

I've got ax-4y=9
-4y=-ax+9
y=(ax/4)-9/4

8x=-11-3y
-3y=8x+11
y=-8/3x-11/3
m= -8/3.

So the slope for the first equation is 3/8.
Placing the slope 3/8 into the equation y=(ax/4)-9/4 is tripping me up.
Any help would be truly appreciated!
Thanks.

 

[Deleted member 4993]

We've gotten a list of problems for my class, and I'm really struggling with this one.

Suppose the line ax-4y=9 is perpendicular to the line 8x=-11-3y. What is the value of a. At what exact point do these two lines intersect?

I know that you have to find the slopes (m) and that if the lines are perpendicular, the slopes will be negative reciprocals.

I've got ax-4y=9
-4y=-ax+9
y=(ax/4)-9/4

8x=-11-3y
-3y=8x+11
y=-8/3x-11/3
m= -8/3.

So the slope for the first equation is 3/8.
Placing the slope 3/8 into the equation y=(ax/4)-9/4 is tripping me up.
Any help would be truly appreciated!
Thanks.

Good you are almost there..

Now from the first equation the slope is "a/4"

so

3/8 = a/4

a = ??

 

[pinkcalculator]

3/2! Thanks so much.
This is a great site, you guys don't just give answers, you explain them.

 

[Denis]

This is a great site, you guys don't just give answers, you explain them.

We help IF you show what you tried, which you did.
Some lazybones come here INSISTING we do their homework (while they play video games, I guess!);
we usually kindly(?) turn them away...