Question

Find the valueof K such that lines where equations are 3kx + 8y = 5 and 6y - 4kx = -1 are perpendicular

Answer 1

Answer:

Step-by-step explanation:

3kx+8y-5=0

8y=5-3kx

y=5/8-3kx/8

Slope= -3/8k  i.e.  m= -3/8k

6y-4kx+1=0

6y=4kx-1

y=4kx/6-1/6

Slope= 4k/6 i.e m1=4k/6

Since lines are perpendicular;

m*m1= -1

-3/8k*4k/6=-1

12k=-48

k= -4

Answer 2

Given:

The equations of two lines 3kx + 8y = 5 and 6y - 4kx = -1 are perpendicular to each other.

To Find:

The value of k.

Solution:

The given problem can be solved using the concepts of straight lines.

1. The given line equations are 3kx + 8y = 5 and 6y - 4kx = -1.

2. Consider two straight lines ax + by + c = 0 and dx + ey + f = 0.  According to the properties of straight lines, these two lines are said to be perpendicular if and only if it satisfies the condition,

=> ad + be = 0.

3. Using the condition mentioned above, the value of k can be determined,

=> 3k x (-4k) + 8 x 6 = 0,

=> -12k² = -48,

=> k² = 4,

=> k = + 2 (OR) k = -2.

Therefore, the value of k such that the lines 3kx + 8y = 5 and 6y - 4kx = -1 are perpendicular is +2 (or) -2.