The centre of the ellipse frac{(x + y - 2)^2} | vedclass

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(B) The given equation of the ellipse is $\frac{(x + y - 2)^2}{9} + \frac{(x - y)^2}{16} = 1$. This is of the form $\frac{X^2}{a^2} + \frac{Y^2}{b^2}

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$(1, 1)$

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(B) The given equation of the ellipse is $\frac{(x + y - 2)^2}{9} + \frac{(x - y)^2}{16} = 1$. This is of the form $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$,where $X = x + y - 2$ and $Y = x - y$. The centre of the ellipse is the point where $X = 0$ and $Y = 0$. Setting $x + y - 2 = 0$ and $x - y = 0$,we get $x = y$. Substituting $x = y$ into the first equation: $x + x - 2 = 0 \implies 2x = 2 \implies x = 1$. Since $x = y$,we have $y = 1$. Thus,the centre is $(1, 1)$.

The centre of the ellipse $\frac{(x + y - 2)^2}{9} + \frac{(x - y)^2}{16} = 1$ is

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