find the value of k such that the lines whose equations are 3x+5ky=7 and 9kx+8y=15 are parallel.

Question

Mathematics

Answer 1

find the value of k such that the lines whose equations are 3x+5ky=7 and 9kx+8y=15 are parallel.

the lines are:

3x + 5ky = 7

9kx + 8y = 15

Convert both these eqns into the form: y = mx + c

In this form, m is the gradient of the line and c is the y-intercept.

In this form, our lines are,

3x + 5ky = 7  -->  y = -(3/5k)x - (7/5k)   i.e. m1 = -(3/5k)

9kx + 8y = 15   -->   y = -(9k/8)x - (15/8)   i.e. m2 = -(9k/8)

The two lines are parallel when they have the same gradient, i.e. m1 = m2

So, 3/5k = 9k/8

24 = 45k^2

8 = 15k^2

k = sqrt(8/15)

Answer 2

The slopes of two parallel lines are the same. To find the slope we ignore the constant term first.

3x+5ky=0 and 9kx+8y=0. Next we divide the x coefficient by the y coefficient and equate them for the two lines: 3/5k=9k/8. Cross-multiply to solve for k: 24=45k^2. So k^2=24/45=8/15 and k=√(8/15)=2√30/15=0.7303. But k can also be -2√30/15.

[If we substitute for k in the two linear equations we get 3x+2y√30/3=7 and 6x√30/5+8y=15.

Or we get 3x-2y√30/3=7 and -6x√30/5+8y=15.]