Two sides of a rhombus are along the lines x | vedclass

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(A) The equations of the lines containing two sides of the rhombus are $L_1: x - y + 1 = 0$ and $L_2: 7x - y - 5 = 0$. The intersection point of these

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$\left( \frac{1}{3}, - \frac{8}{3} \right)$

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(A) The equations of the lines containing two sides of the rhombus are $L_1: x - y + 1 = 0$ and $L_2: 7x - y - 5 = 0$. The intersection point of these two lines is the vertex $A$. Solving $x - y = -1$ and $7x - y = 5$ gives $6x = 6$,so $x = 1$ and $y = 2$. Thus,$A = (1, 2)$. The diagonals of a rhombus bisect the angles between the sides. The equations of the angle bisectors are given by $\frac{x - y + 1}{\sqrt{1^2 + (-1)^2}} = \pm \frac{7x - y - 5}{\sqrt{7^2 + (-1)^2}}$. This simplifies to $5(x - y + 1) = \pm \sqrt{2}(7x - y - 5) / \sqrt{2}$,which is $5(x - y + 1) = \pm (7x - y - 5)$. Case $1$: $5x - 5y + 5 = 7x - y - 5$ $\Rightarrow 2x + 4y - 10 = 0$ $\Rightarrow x + 2y - 5 = 0$. Case $2$: $5x - 5y + 5 = -7x + y + 5$ $\Rightarrow 12x - 6y = 0$ $\Rightarrow 2x - y = 0$. Since the diagonals pass through the intersection point $(-1, -2)$,the diagonal equations are $x + 2y + 5 = 0$ and $2x - y = 0$. The vertex $A(1, 2)$ lies on the line $2x - y = 0$. The other diagonal is $x + 2y + 5 = 0$. The vertex $C$ is the reflection of $A(1, 2)$ across the diagonal $x + 2y + 5 = 0$. Using the reflection formula $\frac{x - 1}{1} = \frac{y - 2}{2} = -2 \frac{1(1) + 2(2) + 5}{1^2 + 2^2} = -2 \frac{10}{5} = -4$. Thus,$x = 1 - 4 = -3$ and $y = 2 - 8 = -6$. The other vertices are found by intersecting the lines $x - y + 1 = 0$ and $7x - y - 5 = 0$ with the diagonals. Checking the options,the point $\left( \frac{1}{3}, - \frac{8}{3} \right)$ satisfies $x + 2y + 5 = 0$ since $\frac{1}{3} + 2(-\frac{8}{3}) + 5 = \frac{1 - 16 + 15}{3} = 0$.

Two sides of a rhombus are along the lines $x - y + 1 = 0$ and $7x - y - 5 = 0$. If its diagonals intersect at $(-1, -2)$,then which one of the following is a vertex of this rhombus?

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