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

A
2
B
5/2
C
7/4
D
7/2

Solution:

Let coordinates of A be $(0, a)$

The diagonals intersect at $P (1, 2)$

We know that the diagonals will be parallel to the

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

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

$\Rightarrow 5 x - 5 y + 10 = \pm (7 x - y + 3)$

$\Rightarrow 2 x + 4 y - 7 = 0$ and $12 x - 6 y + 13 = 0$

$\Rightarrow m_{1} = - 1 / 2$ and $m_{2} = 2$

(where $m_{1}$ and $m_{2}$ are the slopes of the given two lines) Let one diagonal be parallel to $2 x + 4 y - 7 = 0$ and other be parallel to $12 x - 6 y + 13 = 0$

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

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

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

But $a \neq 0$ $∴$ Thus, ordinate of A is 5/2.

The sides of a rhombus ABCD are parallel to the lines, $x - y + 2 = 0$ and $7 x - y + 3 = 0$ . 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

Solution:

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