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(A) The sides of the rhombus are parallel to $x+y+c_1=0$ and $7x-y+c_2=0$. Since the centre is $(1,3)$,the diagonals bisect the angles between the sid

Vedclass · Accepted Answer

$\frac{18}{5}$

Vedclass · Answer

(A) The sides of the rhombus are parallel to $x+y+c_1=0$ and $7x-y+c_2=0$. Since the centre is $(1,3)$,the diagonals bisect the angles between the sides. The lines passing through the centre $(1,3)$ parallel to the sides are $x+y-4=0$ and $7x-y-4=0$. The diagonals of a rhombus are perpendicular bisectors of each other. The slopes of the sides are $m_1 = -1$ and $m_2 = 7$. The slopes of the diagonals are $m_d = \frac{1-m_1m_2 \pm \sqrt{(1-m_1^2)(1-m_2^2)}}{m_1+m_2}$ is not applicable here; rather,the diagonals bisect the angles between the sides. The angle bisectors of $x+y-4=0$ and $7x-y-4=0$ are $\frac{x+y-4}{\sqrt{2}} = \pm \frac{7x-y-4}{\sqrt{50}}$. Simplifying,$5(x+y-4) = \pm (7x-y-4)$. Case $1$: $5x+5y-20 = 7x-y-4 \implies 2x-6y+16=0 \implies x-3y+8=0$. Case $2$: $5x+5y-20 = -7x+y+4 \implies 12x+4y-24=0 \implies 3x+y-6=0$. Vertex $A(\alpha, \beta)$ lies on $15x-5y=6$ and on one of the diagonals. Solving $3x+y=6$ and $15x-5y=6$: $15x+5y=30$ and $15x-5y=6 \implies 30x=36 \implies x=1.2, y=2.4$. $\alpha+\beta = 3.6 = \frac{18}{5}$. Solving $x-3y=-8$ and $15x-5y=6$: $5x-15y=-40$ and $45x-15y=18 \implies 40x=58 \implies x=1.45, y=3.15$. $\alpha+\beta = 4.6 = \frac{23}{5}$ (not in options). Thus,the correct value is $\frac{18}{5}$.

Two non-parallel sides of a rhombus are parallel to the lines $x+y-1=0$ and $7x-y-5=0$. If $(1,3)$ is the centre of the rhombus and one of its vertices $A(\alpha, \beta)$ lies on $15x-5y=6$,then one of the possible values of $(\alpha+\beta)$ is

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