Find P and k if the equation
`px^(2)-8xy+3y^(2)+14x+2y+k=0`
represents a pair of perpendicular lines.

Explanation

The given equation represents a pair of lines perpendicular to each other, we have
(Coeff.of `x^(2)` ) + (Coeff.of `y^(2)`) = 0
`:.P+3=0`
`:.P=-3`
Substituting `P=-3` in the given equation, we get
`-3x^(2)-8xy+3y^(2)+14x+2y+k=0`
On comparing above equation with equation
`ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`, we get
`a=-3, h=-4, b=3, g=7, f=1, k=c`

`:.D=|(a,h,g),(h,b,f),(g,f,c)|=|(-3,-4,7),(-4,3,1),(7,1,k)|`
`=-3(3k-1)+4(-4k-7)+7(-4-21)`
`=-9k+3-16k-28-175`
`=-25k-200`
`:'` The given equation represents a pair of lines, we have `D=0`
`:.-25k-200=0`
`:.k=-8`
`:.P=-3` and `k=-8`.