Question

Two sides of the rhombus PQRS are parallel to the lines $7x^2 - 8xy + y^2 + 17x - 5y + 6 = 0$. If the diagonals of rhombus intersect at point (1, 2) and P lies on Y-axis, then

• A. co-ordiantes of P can be (0,0)
• B. co-ordiantes of P can be $\left(0, \frac{5}{2}\right)$
• C. sum of slope of diagonals of rhombus is $\frac{3}{2}$
• D. sum of slope of diagonals of rhombus is $\frac{5}{2}$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The given equation $7x^2 - 8xy + y^2 + 17x - 5y + 6 = 0$ factors into two lines. The quadratic part $7x^2 - 8xy + y^2$ can be factored as $(y-x)(y-7x)$. Thus, the lines are $y=x$ and $y=7x$. The slopes of these lines are $m_1=1$ and $m_2=7$. These are the slopes of the adjacent sides of the rhombus.

The diagonals of a rhombus are the angle bisectors of the sides. The angle bisectors of $y-x=0$ and $y-7x=0$ are found using the formula: $\frac{x-y}{\sqrt{1^2+(-1)^2}} = \pm \frac{7x-y}{\sqrt{7^2+(-1)^2}}$ $\frac{x-y}{\sqrt{2}} = \pm \frac{7x-y}{\sqrt{50}} = \pm \frac{7x-y}{5\sqrt{2}}$ $5(x-y) = \pm (7x-y)$

Case 1: $5x-5y = 7x-y \implies 2x+4y = 0 \implies y = -\frac{1}{2}x$. The slope is $m_{d1} = -\frac{1}{2}$. Case 2: $5x-5y = -(7x-y) \implies 5x-5y = -7x+y \implies 12x-6y = 0 \implies y = 2x$. The slope is $m_{d2} = 2$.

The slopes of the diagonals are $2$ and $-\frac{1}{2}$. The sum of the slopes is $2 + (-\frac{1}{2}) = \frac{3}{2}$. This confirms option C.

The diagonals intersect at (1, 2). The equations of the diagonals are: Diagonal 1 (slope 2): $y - 2 = 2(x - 1) \implies y = 2x$. Diagonal 2 (slope $-\frac{1}{2}$): $y - 2 = -\frac{1}{2}(x - 1) \implies 2y - 4 = -x + 1 \implies x + 2y - 5 = 0$.

Vertex P lies on the Y-axis, so its coordinates are $(0, y_P)$. P must lie on one of the diagonals. If P lies on $y=2x$: $y_P = 2(0) \implies y_P = 0$. So P can be (0,0). This confirms option A. If P lies on $x+2y-5=0$: $0 + 2y_P - 5 = 0 \implies 2y_P = 5 \implies y_P = \frac{5}{2}$. So P can be $(0, \frac{5}{2})$. This confirms option B.