Question

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.

Answer 1

A being on y axis may be chosen as(0,a) .The diagonals intersect at P(1,2) .
Again we know that the diagonals will be parallel to the angle bisectors of the two sides  y=x+2    and y=7x+3 
i,e [tex] \frac{x-y+2}{ \sqrt{2} } =+- \frac{7x-y+3}{5 \sqrt{2} } [/tex]
5x-y+10=+-(7x-y+3)
2x+4y-7=0 and 12x-6y+13=0 
$m_{1}$=-1/2 and m$m_{2} =2$
Let diagonal d be parallel to 2x+4y-7=0 and digonals D be parallel to 12x-6y+13=0 .The vertex A could be on any of the two diagonals .Hence slope of AP is either -1/2 or 2 
⇒$\frac{2-a}{1-0} =2or -1/2$
⇒a=0 or 5/2 

Answer 2

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 m1.m2=-1

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