Question

6. Find the values of p for which the straight lines 8px+(2-3p)y+1=0 and
px + 8y-7 = 0 are perpendicular to each other

Answer 1

If two lines are perpendicular,then product of their slopes will be equal to -1.

     Slope (m) = - coefficient of x/coefficient of y

Slope of the first line 8px + (2-3p)y + 1 = 0

                  m1 = -8p/(2-3p)

Slope of the second line px + 8y + 7 = 0

                   m2 = -p/8

             m1 x m2 = -1

          [-8p/(2-3p)] x [-p/8] = -1

              [(8 p²)/8(2-3p)] = -1

                        p²/(2-3p) = -1

                        p² = -1(2-3 p)

                        p² = -2 + 3 p

                        p²- 3 p + 2 = 0

                  (p - 2) (p - 1) = 0

          p - 2 = 0         p - 1 = 0

            p = 2             p = 1

Answer 2

Solution :

i ) Compare 8px + ( 2 -3p)y+1 = 0

with ax + by + c = 0 ,

a = 8p , b = 2 - 3p , c = 1

slope ( m1 ) = -a/b

=> m1 = - ( 8p )/( 2 - 3p ) ---( 1 )

ii ) Slope of a line px + 8y - 7 = 0

slope ( m2 ) = - a/b

=> m2 = - ( p/8 ) -----( 2 )

According to the problem given ,

m1 × m2 = -1

=> [ -8p/( 2 - 3p ) ] × ( -p/8 ) = -1

=> p²/( 2 - 3p ) = -1

=> p² = -( 2 - 3p )

=> p² + 2 - 3p = 0

=> p² - 3p + 2 = 0

=> p² - p - 2p + 2 = 0

=> p( p - 1 ) - 2( p - 1 ) = 0

=> ( p - 1 )( p - 2 ) = 0

=> p - 1 = 0 or p-2 = 0

p = 1 or p = 2

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