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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 P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the ordinate of A is :

Option 1) 5/2

Option 2) 7/4

Option 3) 2

Option 4) 7/2

Answers (1)

best_answer

As we learned,

Condition for parallel lines -

$m_{1}= m_{2}$

- wherein

Here $m_{1},m_{2}$ are the slope of two lines

and

Equation of angle bisectors -

$\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}} =\frac{\pm a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}$

- wherein

Angle bisectors of the lines $a_{1}x+b_{1}y+c_{1}=0$ and $a_{2}x+b_{2}y+c_{2}=0$

Let co-ordinate of A = (0,a)

Equation of parallel lines are

x - y + 2 =0 and 7x -y + 3 = 0

Diagonals are $\parallel$ to $\angle bisectors$ i.e.

$\frac{x-y+2}{\sqrt{2}}=\pm \left ( \frac{7x-y+3}{5\sqrt{2}} \right )$

i.e. $L_{1} : \: 2x+4y-7=0$

$L_{2} : \: 12x-6y+13=0$

$m_{1}=\frac{-1}{2}$ and $m_{2}=2$

Slope of A(0,C) to P(1,2) is $\frac{2-C}{1}=\frac{-1}{2}\: \Rightarrow \: C=\frac{5}{2}$

Option 1)

$\frac{5}{2}$

Option 2)

$\frac{7}{4}$

Option 3)

2

Option 4)

$\frac{7}{2}$

Posted by

Himanshu

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