Question

solve the equations

12p = 96
​

Answer 1

Answer:

Answer:

< html > < head > < style > body { background:linear-gradient(100deg,purple,blue, green,lime,yellow); } div{ box-shadow:0px 0px 20px #000; width:250px; height:auto; position:relative; margin:100px auto; background:#fff; border-radius:10px; font-family:Verdana; color:#333; background-color:black; } p{ padding:50px; padding-top:10px; text-align:justify; color:gold; } h1{ background:#b71540; display:inline-block; color:#fff; padding:5px 25px; position:relative; left:-30px; border-bottom-left-radius:5px; } h1:before{ position:absolute; content:''; width:50px; height:50px; background:#b71540; top:20px; left:10px; transform:rotate(45deg); z-index:-1; } < /style > < /head > < body > < br > 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 < br > < strong > Mr Shivam Gurjar < /strong > < /p > < /div > < /body > < /html >