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

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Solution

given straight lines are -3kx + y -4 and (2k-1) x - (8k-1) y - 6=0

these lines are perpendicular if a1.a2 +b1.b2=0
(-3k).(2k-1) + 1 .-(8k-1) =0
-6k^2 +3k-8k+1=0
6k^2 +5k-1=0
(k+1) (6k-1)=0

k=-1 or 1/6

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k

y - 3 k x + 4 = 0

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

k

y - 3 x + 4 = 0

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

x + (k + 1) y = 5

(k + 1) x + 9 y = 8 k - 1

x + (k + 1) y = 5, (k + 1) x + 9 y = (8 k - 1) .

Find the value of k so that the following system of equation has no solution

3x+y=1 (2k-1)x+(k-1)y=o

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