The abscissae of two points P and Q are the r | vedclass

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Question

(A) Let the coordinates of points $P$ and $Q$ be $(x_1, y_1)$ and $(x_2, y_2)$ respectively.Given that $x_1$ and $x_2$ are the roots of $2x^2 + 4x - 7

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$(-1, 2)$

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(A) Let the coordinates of points $P$ and $Q$ be $(x_1, y_1)$ and $(x_2, y_2)$ respectively.Given that $x_1$ and $x_2$ are the roots of $2x^2 + 4x - 7 = 0$,the sum of the roots is $x_1 + x_2 = -\frac{4}{2} = -2$.Given that $y_1$ and $y_2$ are the roots of $3x^2 - 12x - 1 = 0$,the sum of the roots is $y_1 + y_2 = -\frac{-12}{3} = 4$.The centre of the circle with $PQ$ as a diameter is the midpoint of $PQ$,which is given by $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.Substituting the values,we get $(\frac{-2}{2}, \frac{4}{2}) = (-1, 2)$.Thus,the correct option is $A$.

The abscissae of two points $P$ and $Q$ are the roots of the equation $2x^2 + 4x - 7 = 0$ and their ordinates are the roots of the equation $3x^2 - 12x - 1 = 0$. Then the centre of the circle with $PQ$ as a diameter is

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