Question

Two sides of a rhombus lying in the first quadrant are given by `3x-4y=0a n d12 x-5y=0` . If the length of the longer diagonal is 12, then find the equations of the other two sides of the rhombus.

Answer 1

OB is along the acute angle bisector of the given lines: 3x-4y=0 and 12x-5y =0.
The equation of acute angle bisector is
`(3x-4y)/(5) = -(12-5y)/(13)`
or 9x-7y=0
`therefore "Slope of OB" = tan theta = (9)/(7)`
Now, OB= 12.
Therefore, the coordinates of B are
`(12 "cos" theta, 12 "sin" theta) -= (12(7)/(sqrt(130)), 12(9)/(sqrt(130))) -= ((84)/(sqrt(130)), (108)/(sqrt(130)))`
AB is passing through B and is parallel to OC.
Therefore, the equation of AB is
`y-(108)/(sqrt(130)) = (12)/(5)(x-(84)/(sqrt(130)))`
BC is passing through B and is parallel to OA.
Therefore, the equation of BC is `y-(108)/(sqrt(130)) = (3)/(4)(x-(84)/(sqrt(130)))`