Question

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?

Answer 1

As we learned,

Condition for parallel lines -

$m_1=m_2$ ------ wherein $m_1,m_2$ are the slope of two lines

Equation of angle bisectors -

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{\pm a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$

(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 $A=(0,a)$

Equations of parallel lines, given, are,

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

Diagonals are parallel to the angular bisectors i.e.

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

i.e., $L_1:2x+4y-7=0$ and $L_2:12x-6y+13=0$

$m_1=\frac{-1}{2}$ and $m_2=2$

Slope of $A(0,C)$ to $P(1,2)$ is

$\frac{2-C}{1}=\frac{-1}{2}⟹C=\frac{5}{2}$

$\therefore A(0,C)=(0,\frac{5}{2})$