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Question

Find $$p$$ and $$q$$ if equation $$p{x^2} - 8xy + 3{y^2} + 14x + 2y + q = 5$$, represents pair of perpendicular lines.


Solution

Given, $$ px^{2}-8xy+3y^{2}+14x+2y+q=5$$...(i) represents pair of 
perpendicular lines.
equation (i) is of the form
$$ ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$$
$$ \Rightarrow  a=p, h=4, b=3, g=7, f=1, c=q-5.$$
since the equation (i) represents pair of perpendicular lines
$$a+b=0$$
$$ \Rightarrow p+3=0$$
$$ \Rightarrow p= -3. ...(ii)$$
also eqn (i) represents a pair of straight lines
$$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$$
$$ \Rightarrow \begin{vmatrix} p & -4 & 7 \\ -4 & 3 & 1 \\ 7 & 1 & q-5 \end{vmatrix} = 0$$
$$ \Rightarrow p(3q - 15 - 1 ) + 4 (-4q + 20 - 7) + 7 (-4-21)=0$$
$$ \Rightarrow (-3)(3q-16) + 4(-4q+13) - 175=0$$ [from (ii)P=-3]
$$ \Rightarrow -9q + 48 - 16q + 52 - 175 = 0$$
$$ \Rightarrow -25q-75=0$$
$$ \Rightarrow -25q=75$$
$$ \therefore q=-3$$
Hence $$p = -3, q = -3.$$

Mathematics

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