Question

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

• A. 18/5
• B. 12/5
• C. 37/5
• D. 39/5

Hint:

A key property of a rhombus is that its diagonals bisect the angles between the sides. This means the diagonals will have the same direction as the angle bisectors of the lines representing the sides.

Answer 1

The Correct Option is A

Step 1: Find the directions of the diagonals.
The diagonals of a rhombus are the angle bisectors of the adjacent sides. The sides of the rhombus are parallel to the lines $L_1: x+y-1=0$ and $L_2: 7x-y-5=0$. Therefore, the diagonals are parallel to the angle bisectors of $L_1$ and $L_2$. The equations of the angle bisectors are given by: \[ \frac{x+y-1}{\sqrt{1^2+1^2}} = \pm \frac{7x-y-5}{\sqrt{7^2+(-1)^2}} \implies \frac{x+y-1}{\sqrt{2}} = \pm \frac{7x-y-5}{5\sqrt{2}}. \] $5(x+y-1) = \pm (7x-y-5)$. Bisector 1 (+): $5x+5y-5 = 7x-y-5 \implies 2x-6y=0 \implies x-3y=0$. Bisector 2 (-): $5x+5y-5 = -7x+y+5 \implies 12x+4y-10=0 \implies 6x+2y-5=0$.

Step 2: Find the equations of the diagonals.
The diagonals must pass through the center of the rhombus, which is $(1,3)$. Diagonal $D_1$ is parallel to $x-3y=0$, so its equation is $(x-1)-3(y-3)=0 \implies x-3y+8=0$. Diagonal $D_2$ is parallel to $6x+2y-5=0$, so its equation is $6(x-1)+2(y-3)=0 \implies 3x+y-6=0$.

Step 3: Find the coordinates of the vertex A.
The vertex $A(\alpha, \beta)$ must lie on one of the diagonals. It also lies on the line $15x-5y=6$. So, we must find the intersection of this line with each diagonal. Case 1: A is the intersection of $x-3y+8=0$ and $15x-5y=6$. From the first equation, $x=3y-8$. Substituting into the second: $15(3y-8)-5y=6 \implies 45y-120-5y=6 \implies 40y=126 \implies y=63/20$. Then $x = 3(63/20)-8 = 189/20 - 160/20 = 29/20$. In this case, $\alpha+\beta = 29/20 + 63/20 = 92/20 = 23/5$. This is not an option. Case 2: A is the intersection of $3x+y-6=0$ and $15x-5y=6$. From the first equation, $y=6-3x$. Substituting into the second: $15x-5(6-3x)=6 \implies 15x-30+15x=6 \implies 30x=36 \implies x=36/30=6/5$. Then $y = 6-3(6/5) = 6-18/5 = 12/5$. So, the vertex is $A(\alpha, \beta) = (6/5, 12/5)$.

Step 4: Calculate the required value.
For the vertex found in Case 2, we calculate the sum $\alpha+\beta$. \[ \alpha+\beta = \frac{6}{5} + \frac{12}{5} = \frac{18}{5}. \] This value matches option (A). \[ \boxed{\alpha+\beta = \frac{18}{5}}. \]