For what values of K ,so that the line L:y=(2k-4)x+6 ,is A)parallel toL2:y-8x=10 B) Perpendicular to L2:y-8x=?

A) To find the values of k for which the line L is parallel to L2, we need to compare the slopes of the two lines. The slope of L is given by the coefficient of x in the equation, which is 2k-4. The slope of L2 is 8.

For two lines to be parallel, their slopes must be equal. Therefore, we can set the slope of L equal to the slope of L2 and solve for k:

2k-4 = 8

Adding 4 to both sides:

2k = 12

Dividing both sides by 2:

k = 6

So, the line L:y=(2k-4)x+6 is parallel to L2:y-8x=10 when k = 6.

B) To find the values of k for which the line L is perpendicular to L2, we need to compare the slopes of the two lines. The slope of L is given by the coefficient of x in the equation, which is 2k-4. The slope of L2 is 8.

For two lines to be perpendicular, the product of their slopes must be -1. Therefore, we can set the product of the slopes equal to -1 and solve for k:

(2k-4)(8) = -1

Expanding and simplifying:

16k - 32 = -1

Adding 32 to both sides:

16k = 31

Dividing both sides by 16:

k = 31/16

So, the line L:y=(2k-4)x+6 is perpendicular to L2:y-8x=10 when k = 31/16.