Question

Let ABCD is a rhombus in which a sides AB and AD are given by the lines
x – y – 2 = 0 and 6x – y + 3 = 0 respectively. Diagonals AC & BD of rhombus ABCD intersect at point P(1, 2)

i)Equation of CD can be
ii)Equation of other diagonal BD is

Answer 1

Answer:

Let the coordinate of A be (0,α).

Since the sides AB and AD are parallel to the lines y=x+2 and y=7x+3 respectively.

Therefore the diagonal AC is parallel to the bisector of the angle between these two lines.

The equation of the bisectors are given by, 

2x−y+2=±507x−y+3⇒5(x−y+2)=±(7x−y+3)

⇒2x+4y−7=0 and 12x−6y+13=0

Thus, the diagonals of the rhombus are parallel to the lines 

2x+4y−7=0 and 12x−6y+14=0

Therefore slope of AE=−42 or 612

⇒1−02−α=−21 or 1−02−α=2

⇒α=25 or α=0

I hope it will be help you