Question

 Find 'p' and 'q' if the equation px² - 8xy +3y² +14 x +2y +q = 0 represents a pair of perpendicular lines.

Answer 1

Given general equation is px² - 8xy +3y² +14 x +2y +q = 0

a = p , h = -4 , b = 3 , g = 7, f = 1 , c = q

Since the equation represents perpendicular lines 

a +b =0

p + 3 = 0

∴ p = -3 

∴ a = p = -3

Since the equation represents a pair of lines.

`abs[("a","h" ,"g"),("h","b","f"),("g" ,"f" , "c")] = 0`

`abs[(-3,-4,7),(-4,3,1),(7,1,"q")] = 0`

∴ -3(3q-1)+4(-4q-7)+7(-4-21)=0

∴ -9q +3 -16q - 28 - 28 - 147 = 0

∴ -25q + 3 - 28 - 175 = 0

∴ -25q + 3 - 203 = 0

∴ -25q - 200 = 0

∴ -25q = 200

∴ q = `200/-25`

∴ q = -8

∴ p = -3 , q = -8