## Find slope of 2 perpendicular lines

1. The problem statement, all variables and given/known data
Find the values of k such that lines 3kx+8y = 5 and 6y -4kx = -1 are perpendicular.

I don't need the answer, but a push in a right direction. However, feel free to solve the equation :)

2. Relevant equations
1. Slope of a line is m = [y2-y1]/[x2-x1] where we have 2 points on the line
P1(x1,y1) and P2(x2,y2)

2. Product of slope of 2 perpendicular lines is -1
let's say m1 is slope of line 1 and m2 is slope of line 2 then
m1 = -(1/m2)

3. The attempt at a solution

My attempts to resolve this equation are just confusing. I know that we have too many variables and we need to get rid of some of them.

I also know that at one point both equations are equal to each other (intersection point)

so...
3kx+8y = 5 and 6y -4kx = -1 are...

1. k = (5-8y)/3x and 2. k = (-1 -6y)/ -4x

if k = -(1/k) then

(5-8y)/3x = -((-4x)/(-1-6y))

(5-8y)(-1-6y) =(4x)*(3x)

-5 -30y +8y +48y^2 = 12x^2

and here I get stuck! What to do then? and am I taking a right path to solve this problem?

Thank you very much for any help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
 Mentor I think you're making it more difficult than it needs to be! You are correct that, if two lines are perpendicular then the product of their gradients is equal to -1. Try writing both equations in the form y=mx+c, where m is the gradient of the curve. Then you can envoke your relation m1m2=-1 and solve for k.
 Cristo: thank you for your reply. It seems to me no matter how much I play with these 2 equations I always have either too many variables or get stuck with x and y square. I forgot to mention that k of line 1 isn't necessarily equals to k of line 2. So I don't see how writting them both in for y =mx +c will help me solve it. 1. y = (5-3kx)/8 2. y =(-1+4kx)/6 I'm still stuck with x being unknown and k of line 1 and k of line 2.

Mentor

## Find slope of 2 perpendicular lines

Ok, since the k's are different, write j in the first, and k in the second equation. Plug these into the formula m1m2=-1 and you will arrive at a relation between j and k.

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cristo, I would not interpret this as two different values for k. The problem is asking for a value for k, in both equations, such that the two lines are perpendicular

 Quote by phoenix20_06 1. The problem statement, all variables and given/known data Find the values of k such that lines 3kx+8y = 5 and 6y -4kx = -1 are perpendicular. I don't need the answer, but a push in a right direction. However, feel free to solve the equation :) 2. Relevant equations 1. Slope of a line is m = [y2-y1]/[x2-x1] where we have 2 points on the line P1(x1,y1) and P2(x2,y2)
A more "relevant" equation is this: if y= mx+ b, then the slope is m. Can you solve 3kx+ 8y= 5 and 6y- 4kx= -1 so you have each in that form and can write the slope as a function of k only? Then use the fact that the product of the slopes must be -1.

 2. Product of slope of 2 perpendicular lines is -1 let's say m1 is slope of line 1 and m2 is slope of line 2 then m1 = -(1/m2) 3. The attempt at a solution My attempts to resolve this equation are just confusing. I know that we have too many variables and we need to get rid of some of them. I also know that at one point both equations are equal to each other (intersection point) so... 3kx+8y = 5 and 6y -4kx = -1 are... 1. k = (5-8y)/3x and 2. k = (-1 -6y)/ -4x if k = -(1/k) then (5-8y)/3x = -((-4x)/(-1-6y)) (5-8y)(-1-6y) =(4x)*(3x) -5 -30y +8y +48y^2 = 12x^2 and here I get stuck! What to do then? and am I taking a right path to solve this problem? Thank you very much for any help!

Mentor
 Quote by HallsofIvy cristo, I would not interpret this as two different values for k. The problem is asking for a value for k, in both equations, such that the two lines are perpendicular
That's what I thought originally, but when the OP said:
 I forgot to mention that k of line 1 isn't necessarily equals to k of line 2.
I thought he had been explicitly told otherwise!
 3kx+8y = 5 and 6y -4kx = -1 are perpendicular. slope of the first one can be obtained from y = 5/8 - 3k/8x and slope of the second one can be obtained from y = -1/6 + (2/3k)x from the first one the slope is -3k/8 and from the second one 2k/3 since the product of slopes is -1 you have -6k^2/24 = -1 k^2/4 = 1 k^2 = 4 Check this, maybe I've messed something up, but according to this k is 2 and -2.

 Quote by phoenix20_06 I forgot to mention that k of line 1 isn't necessarily equals to k of line 2. So I don't see how writting them both in for y =mx +c will help me solve it.
y = mx + c can be used if they are the same. Are you trying to say that you need to find an equation for two different 'k's? e.g. k1 + k2 = c, where c is some constant. More like a linear programming question.

 Quote by phoenix20_06 1. y = (5-3kx)/8 2. y =(-1+4kx)/6 I'm still stuck with x being unknown and k of line 1 and k of line 2.
You do not need to know x and at no stage do you need to. You are writing the equations in the form y = mx + c because you want to know what m is. x makes no difference to the gradient. Knowing the two 'm's from the two, you can use m1 x m2 = -1, as you stated. However........

 Quote by *best&sweetest* Check this, maybe I've messed something up, but according to this k is 2 and -2.
I concour with *best&sweetest*, although I do need to remind you that the person that asked the question is suppose to work through the steps and understanding to produce the solution.

Cheers,

 Quote by phoenix20_06 I forgot to mention that k of line 1 isn't necessarily equals to k of line 2. So I don't see how writting them both in for y =mx +c will help me solve it.
As the solution for k1 = k2 has been given, I am going to go through a more general solution, which is more or less the same as post #7. All you have to do is replace the 'k's, one with k1 and the other with k2. This will produce k1 x k2 = 4, which is obvious from the posts before. So find any two values that satisfy this equation and you will have all of the solutions for k, assuming they differ.

 Mentor Are we all ignoring the homework help rules today? Please do not post full solutions. Instead try to help the OP to get the answer for himself!