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If the combined equation of the diagonals of the square formed by the pairs of lines $x y+4 x-5 y-20=0$ and $x y-5 x+4 y-20=0$ is $x^2-y^2-k x+l y=0$, then $k+l=$.
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Verified Answer
The correct answer is:
2
$$
\text { We have, }
$$
(i)
$$
\begin{aligned}
x y+4 x-5 y-20 & =0 \\
(x-5)(y+4) & =0
\end{aligned}
$$
(ii)
$$
\begin{aligned}
x y-5 x+4 y-20 & =0 \\
(x+4)(y-5) & =0
\end{aligned}
$$
Equation of diagonal $A C$
$$
\begin{aligned}
& y+4=x+4 \\
& \Rightarrow \quad x-y=0 \\
&
\end{aligned}
$$
Equation diagonal $B D$
$$
\begin{array}{rlrl}
y+4 & =-(x-5) \\
\Rightarrow & x+y-1 & =0
\end{array}
$$
Combined equation of diagonal are
$$
\begin{array}{rlrl}
(x-y)(x+y-1) & =0 \\
x^2-y^2-x+y & =0 \\
\therefore & k & =1, l=1
\end{array}
$$
Hence, $k+l=1+1=2$
\text { We have, }
$$
(i)
$$
\begin{aligned}
x y+4 x-5 y-20 & =0 \\
(x-5)(y+4) & =0
\end{aligned}
$$
(ii)
$$
\begin{aligned}
x y-5 x+4 y-20 & =0 \\
(x+4)(y-5) & =0
\end{aligned}
$$
Equation of diagonal $A C$
$$
\begin{aligned}
& y+4=x+4 \\
& \Rightarrow \quad x-y=0 \\
&
\end{aligned}
$$
Equation diagonal $B D$
$$
\begin{array}{rlrl}
y+4 & =-(x-5) \\
\Rightarrow & x+y-1 & =0
\end{array}
$$
Combined equation of diagonal are
$$
\begin{array}{rlrl}
(x-y)(x+y-1) & =0 \\
x^2-y^2-x+y & =0 \\
\therefore & k & =1, l=1
\end{array}
$$
Hence, $k+l=1+1=2$
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