Question

Find the value of k if the straight lines y-3kx+4=0,(2k-1)x-(8k-2)y-6=0 are perpendicular

Answer 1

Step-by-step explanation:

y-3kx+4=0 ......(1)

y = 3kx - 4

If we compare it with standard form y = mx + c where m is slope we get

slope of line (1) is m1 = 3k

And

Similarly for other equation

(2k - 1)x - (8k - 1)y - 6 = 0

⇒ ( 8k - 1 )y = (2k - 1)x - 6

⇒ y = ( 2k-1 )x / ( 8k - 1 ) - ( 6/(8k-1) )

If we compare it with standard form y = mx + c where m is slope we get

slope of line (2) is m2 = ( 2k-1 ) / ( 8k - 1 )

IF line 1 and line 2 is perpendicular to each other then

m1 = -1/m2

Putting the values we get

3k = - ( 8k - 1 ) / ( 2k-1 )

multiplying by 2k -1 on both sides we get

3k(2k - 1) = - (8k - 1)

6k² - 3k = - 8k + 1

6k² - 3k + 8k - 1 = 0

6k² + 5k - 1 = 0

6k² + 6k - k -1 = 0

6k(k + 1) -1(k + 1) = 0

(k + 1)(6k - 1) = 0

⇒ k + 1 = 0 or 6k -1 = 0

k = -1 or k = 1/6

So for k = -1 or k = 1/6 lines are perpendicular to each other