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

Explanation

Given , `px^(2) - 8xy + 3y^(2) + 14x + 2y + q = 0 ` represents
pair of perpendicular lines .
Compaing it with ` ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c = 0 `
` therefore ` we get , a = p , 2h = - 8 (h = - 4), b = 3 , 2g = 14 ,
(g = 7) 2f = 2, (f = 1) and c = q
The equation represents a pair of lines
` therefore abc + 2fgh - af^(2) - bg^(2) - ch^(2) = 0 `
Now put the values
` therefore 3pq + 2xx 1 xx 7 (-4) -p (1)^(2) - 3 (7)^(2) - q (-4)^(2) = 0 `
` therefore 3 pq - 56 - p - 147 - 16q = 0 ` ...(i)
Since , lines are perpendicular
` therefore a + b = 0 `
` therefore p + 3 = 0 `
` therefore p = - 3 `
Put the value of p in equation (i) , we get
` rArr " we get ", - 9q - 65 + 3 - 147 - 16 q = 0 `
` rArr - 25 q - 200 = 0 `
`rArr - 25 q = 200`
`rArr q = - 8 `
Hence , p = - 3 and q = -8` .