The sides of a rhombus ABCD are parallel to the lines,x−y+2=0 and7x−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

# 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  and the vertex A (different from the origin) is on the y-axis, then the ordinate of A is

1. A

2

2. B

5/2

3. C

7/4

4. D

7/2

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### Solution:

Let coordinates of A be

The diagonals intersect at

We know that the diagonals will be parallel to the

angle bisectors of the two sides $y=x+2$ and $y=7x+3$

i.e. $\frac{x-y+2}{\sqrt{2}}=±\frac{7x-y+3}{5\sqrt{2}}$

$⇒5x-5y+10=±\left(7x-y+3\right)$

$⇒2x+4y-7=0$ and $12x-6y+13=0$

$⇒{m}_{1}=-1/2$ and ${m}_{2}=2$

(where ${m}_{1}$ and ${m}_{2}$ are the slopes of the given two lines) Let one diagonal be parallel to and other be parallel to $12x-6y+13=0$

The vertex A could be on any of the two diagonals.

Hence, slope of AP is either$-1/2$ or 2.

$⇒\frac{2-a}{1-0}=2$ or$\frac{2-a}{1-0}=-\frac{1}{2}⇒a=0$ or$a=\frac{5}{2}$

But $a\ne 0$       $\therefore$                                                              Thus, ordinate of A is 5/2.

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