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 equations is `px^(2) - 8xy + 3y^(2) + 14x + 2y + q = 0`
It represents a pair of lines perpendicular to each other .
`therefore " "("coeff.of" x^(2)) +("coeff.of" y^(2)) = 0`
`therefore" "p+3 = 0`
`therefore" "p= - 3`
Putting the value of p is given equaiton,
`-3x^(2) - 8xy + 3y^(2) + 14x + 2y + q =0`
Comparing this equation with .
`ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c =0 `, we have
`a = -3, h = - 4, b = 3, g = 7 , f =1 and c = q`
`therefore " "D = |{:(a,h,g),(h,b,f),(g,f,c):}|`
`= |{:(,-3,-4,7),(,-4,3,7),(,7,1,q):}|`
` = -3 (3q -1) + 4(-4q - 7) +7(-4-21)`
` = - 9q + 3 - 16q - 28 - 175`
` = - 25q - 200`
` = - 25(q + 8)`
Since the equation represents a pair of lines,
`therefore" " D = 0`
`rArr " " - 25 (q+8) = 0`
` rArr" " q = - 8`
`rArr" " q = - 8`
Hence, `p = - 3` and `q = - 8`