The midpoint of the diagonal of a rectangle f | vedclass

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(C) The given equations are $x^2+5x-6=0$ and $y^2-8y-20=0$.Solving for $x$:$x^2+6x-x-6=0$ $\Rightarrow (x+6)(x-1)=0$ $\Rightarrow x_1=-6, x_2=1$.Solvi

Vedclass · Accepted Answer

$\left(\frac{-5}{2}, 4\right)$

Vedclass · Answer

(C) The given equations are $x^2+5x-6=0$ and $y^2-8y-20=0$.Solving for $x$:$x^2+6x-x-6=0$ $\Rightarrow (x+6)(x-1)=0$ $\Rightarrow x_1=-6, x_2=1$.Solving for $y$:$y^2-10y+2y-20=0$ $\Rightarrow (y-10)(y+2)=0$ $\Rightarrow y_1=10, y_2=-2$.The vertices of the rectangle are formed by the intersection of these lines: $(-6, 10), (1, 10), (1, -2), (-6, -2)$.The midpoint of the diagonal is the midpoint of the segment connecting opposite vertices,such as $(-6, 10)$ and $(1, -2)$.Midpoint $= \left(\frac{-6+1}{2}, \frac{10-2}{2}\right) = \left(\frac{-5}{2}, \frac{8}{2}\right) = \left(\frac{-5}{2}, 4\right)$.

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

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