find the values of k,if the straight lines y-3kx+4=0 and (2k-1)x-(8k-1) y-6=0 are perpendicular.
Answers
Answer:
k = -1 or k = 1/6
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
Answer:
K=-1 or k=1/6
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
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