A pair of straight lines x^2 - 8x + 12 = 0 an | vedclass

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(B) Given equations of the lines are $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$.Solving $x^2 - 8x + 12 = 0$ gives $(x - 6)(x - 2) = 0$,so $x = 2$ an

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$(4, 7)$

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(B) Given equations of the lines are $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$.Solving $x^2 - 8x + 12 = 0$ gives $(x - 6)(x - 2) = 0$,so $x = 2$ and $x = 6$.Solving $y^2 - 14y + 45 = 0$ gives $(y - 5)(y - 9) = 0$,so $y = 5$ and $y = 9$.The sides of the square are $x = 2, x = 6, y = 5, y = 9$.The centre of the square is the intersection of the midlines of the parallel sides.The $x$-coordinate of the centre is $\frac{2 + 6}{2} = 4$.The $y$-coordinate of the centre is $\frac{5 + 9}{2} = 7$.Thus,the centre of the inscribed circle is $(4, 7)$.

$A$ pair of straight lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ form a square. The coordinates of the centre of the circle inscribed in the square are

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