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 P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :
Solutions · 1
Official solution
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 =2−1⇒c =25∴Coordinate of A is25