QJEEM18Ma5e38: JEE Main 2018 Mathematics - Class 11 JEEM18Ma5e38 Solution: The sides of a rhombus ABCD are parallel to the lines, x y + 2…

Let the coordinate A be (0, c)

Equations of the given lines are

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

We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3

$\therefore\,\,\,$ equation of angle bisectors is given as :

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

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

$\therefore\,\,\,$ Parallel equations of the diagonals are 2x + 4y $-$ 7 = 0

and 12x $-$ 6y + 13 = 0

$\therefore\,\,\,$ slopes of diagonals are ${{ - 1} \over 2}$ and 2.

Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 $-$ c)

$\therefore\,\,\,$ 2 $-$ c = 2 $\Rightarrow$  c = 0 (not possible)

$\therefore$ $\,\,\,$ 2 $-$ c = ${{ - 1} \over 2}$ $\Rightarrow$ c = ${5 \over 2}$

$\therefore\,\,\,$ Coordinate of A is ${5 \over 2}$ .