Question

The value of K such that the
straight lines y – 3kx + 4 = 0 and
(2k – 1) x – (8k – 1) y – 6 = O are
perpendicular is​

Answer 1

Answer:

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Step-by-step explanation:

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 2

the inventor of the triangle formed by the line 3x+4y-12=0 with the coordinate axes id