Question

Find p and q If the following equation 12 x square - 7 x y + pY square - 25 x + 2qY - 7 equal to zero represent a pair of perpendicular lines

Answer 1

Given equation is

12x

2

+7xy−py

2

−18x+qy+6=0

On comparing with

ax

2

+by

2

+2hxy+2gx+2fy+c=0, we get

a=12,b=−p,h=

2

7

,g=−9,f=

2

q

,c=6

Conditions for pair of lines and pair of perpendiculars are

abc+2fgh−af

2

−bg

2

−ch

2

=0 and a+b=0

⇒12×(−p)(6)+2×(

2

q

)(−9)(

2

7

)−12(

2

q

)

2

−(−p)(−9)

2

−6(

2

7

)

2

=0

and 12−p=0

⇒−72p−

2

63

q−3q

2

+81p−

2

147

=0

and 12−p=0

⇒2q

2

+21q−23=0 and p=12

$\Rightarrow q = 1andp = 12$$.