given lines L1 =(1-a) x-y =1 and L2= 6x+ay =9 find the value of a when L1 and L2 are parallel and L1 and L2 are perpendicular ​

Explanation:

Given lines

L1

( 1 - a) x - y = 1 i)

L2

6x + ay = 9 ii)

Slope of line I) Is

m1 = - ( x-coefficient) / ( y-coefficient)

= - ( 1 - a) / - 1

= 1 - a

Slope of line II) is

m2 = - (x-coefficient) / ( y-coefficient)

= - 6 / a

Now by the question

In 1st case

L1 and L2 are parallel so

m1 = m2

( 1 - a) = - 6 / a

a - a² = - 6

a² - a - 6 = 0

a² - ( 3 - 2 ) a - 6 = 0

a² - 3a + 2a - 6 =0

a ( a - 3 ) + 2 ( a - 3 ) = 0

(a - 3 ) ( a + 2) = 0

Either

a = 3 or a = - 2

Now in second case

L1 and L2 are perpendicular so

m1 * m2 = - 1

1 - a * - 6/ a = - 1

- 6 + 6a = - a

6a + a = 6

7a = 6

a = 6 / 7

The value of a are 3 , -2 and 6/7

Hope it will help :)❤

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