Two sides rhombus ABCD are parallel to the lines

y=x+2 and y=7x+3.if the diagonals of the rhombus intersect at the point(1,2)& the vertex A is on yhe y axis,find the possible coordinates of A.

The answer should be c=0,2.5
Assume A(0,c).

Now opposite sides in rhombus are parallel and all sides are equal length.

Or to check if a quardilateral is rhombus, it is sufficient to show that diagonals are perpendicular bisectors.
Lets assume it is a rhombus, A(0,c). C(2,4-c) as mid-point of AC is (1,2).
Assume line AB is parallel to y=x+2, and AD is parallel to y=7x+3
AB will be y=x+c, and AD will be y=7x+c,

CD will be parallel to AB, if you substitute C in a general line y=x+m, u will get CD : y=x+(2-c),

//ly find BD which will be parallel to AC, BD will be y=7x-(10+c),

Now you have all the line equations, find all points A,B,C,D.
The mid-point of line BD will be (1,2) and BD will be perpendicular to AC,

To prove two lines are perpendicular m_1.m₂=-1

If you put these two conditions, you will get c=0,2.5

Alternative solutions will be to use vectors,
Let A(0,c). Assume side length=d.

AC will be = (1-0) i +(2-c) j,

Line AD can be in 1^st or 2^nd quardent,

Point B = (0 i + c j) + d/rt(2) * (i + j),

Point D = (0 i + c j) (+or-) d/5.rt(2) * (i + 7j),

Mid-point of BD is (1,2),

For first case(+), Mid-point criteria:

d/rt(2) + d/5.rt(2) = 2, d=5.rt(2)/3,

d/rt(2) + 7d/5.rt(2) = 4, d=5.rt(2)/3, same as for x-cordinate :)

BD is perpendicular to AC will give you c=0,

For case(-), Mid-point of BD will be (1,2)

d/rt(2) – d/5.rt(2) = 2, d=5.rt(2)/2,

and 2c + d/rt(2) – 7d/5.rt(2) = 4

Putting d=5.rt(2)/2 you will get, c=2.5,

BD will be perpendicular to AC will give you the same result, ie. c=2.5

using vectors will save you a lot of time.
ATB

The line joining A to P will be the angle bisector of AB and AD. Just assume A(0,c). Slope of AP=2-c..
Now apply formula of tan()=m1-m2/1+m1m2 for the three lines