The midpoint of the diagonal of a rectangle formed by $x^2+5 x-6=0$ and $y^2-8 y-20=0$ is

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \mathrm{x}^2+5 \mathrm{x}-6=0 \Rightarrow \mathrm{x}^2+6 \mathrm{x}-\mathrm{x}-6=0 \ & \Rightarrow(\mathrm{x}-1)(\mathrm{x}+6)=0 \Rightarrow \mathrm{x}_1, \mathrm{x}_2=1,-6 \ & \mathrm{y}^2-8 \mathrm{y}-20=0 \Rightarrow \mathrm{y}^2-10 \mathrm{y}+2 \mathrm{y}-20=0 \ & \Rightarrow(\mathrm{y}-10)(\mathrm{y}+2)=0 \Rightarrow \mathrm{y}_1, \mathrm{y}_2=-2,10 \end{aligned} $ $	herefore$ Midpoint of the diagonal $ =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}ight)=\left(\frac{-5}{2}, 4ight) $

The midpoint of the diagonal of a rectangle formed by $x^2+5 x-6=0$ and $y^2-8 y-20=0$ is

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