Question

Find the value of k so that the line 2x+ky-9=0 may be parallel to 3x-4y+7=0

Answer 1

Answer:

The  value of k = -8/3

Step-by-step explanation:

The two lines represented by the equation

a1x + b1y + c1z = 0 and a2x + b2y + c2 =0

are parallel, then

a1/a2 = b1/b2 ≠ c1/c2

To find the value of k

It is given that, equations of two lines are

2x+ky-9=0  parallel to 3x-4y+7=0

a1 = 2 ,b1 = k , c1 = -9  and a2 = 3 , b2 = -4 and c2 = 7

by the theorem a1/a2 = b1/b2 then,

2/3 = k/-4

k = -8/3

Therefor value of k = -8/3

Answer 2

Answer:

k = - 8/3

Step-by-step explanation:

2x + ky - 9 = 0

ky = -2x + 9

y = - 2/k x + 9/k

Gradient = - 2/k

3x - 4y + 7 = 0

4y = 3x + 7

y = 3/4 x + 7/4

Gradient = 3/4

Since 2x + ky - 9 = 0 and 3x - 4y + 7 = 0 are parallel

⇒ They have the same gradient

Solve k:

-2/k = 3/4

3k = - 8

k = - 8/3

Answer: k = -8/3