The sides of a rhombus ABCD are parallel to t | vedclass

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(D) Let the coordinate of vertex $A$ be $(0, c)$.The equations of the lines parallel to the sides are $x - y + 2 = 0$ and $7x - y + 3 = 0$.The diagona

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$\frac{5}{2}$

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(D) Let the coordinate of vertex $A$ be $(0, c)$.The equations of the lines parallel to the sides are $x - y + 2 = 0$ and $7x - y + 3 = 0$.The diagonals of a rhombus are the angle bisectors of the lines containing the sides.The equations of the angle bisectors are given by $\frac{x - y + 2}{\sqrt{1^2 + (-1)^2}} = \pm \frac{7x - y + 3}{\sqrt{7^2 + (-1)^2}}$.$\frac{x - y + 2}{\sqrt{2}} = \pm \frac{7x - y + 3}{5\sqrt{2}}$.$5x - 5y + 10 = \pm (7x - y + 3)$.Case $1$: $5x - 5y + 10 = 7x - y + 3 \Rightarrow 2x + 4y - 7 = 0$. The slope is $m_1 = -\frac{1}{2}$.Case $2$: $5x - 5y + 10 = -7x + y - 3 \Rightarrow 12x - 6y + 13 = 0$. The slope is $m_2 = 2$.The diagonals pass through $P(1, 2)$ and $A(0, c)$. The slope of the line $AP$ is $\frac{2 - c}{1 - 0} = 2 - c$.If $2 - c = 2$,then $c = 0$,which corresponds to the origin (not allowed).If $2 - c = -\frac{1}{2}$,then $c = 2 + \frac{1}{2} = \frac{5}{2}$.

The sides of a rhombus $ABCD$ are parallel to the lines $x - y + 2 = 0$ and $7x - y + 3 = 0$. If the diagonals of the rhombus intersect at $P(1, 2)$ and the vertex $A$ (different from the origin) is on the $y$-axis,then the ordinate of $A$ is

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