The two sides of rhombus $ABCD$ are parallel to the line $y=x+2$ and $y=3x+7$.
If the diagonals of the Rhombus intersect at point $(1,2)$ and the vertex lies on the $y$-axis then find the possible coordinates of $A$.
I tried this out using the slope Formula $m_1\cdot m_2=-1.$ I wasn't able to solve it further. Please help me out.
Assuming vertex $A$ lies on the $y$ axis and we have a point $(0,y_0)$. Now since the sides are parallel to $y=x+2$ and $y=3x+7$. The equation of the sides would have the same slope respectively, also they would satisfy the point $(0,y_0)$ and will have the same $y$-intercepts because they would intersect at $(0,y_0)$, let say $c$, we have the equation of the sides as:
$$y=x+c\tag{1}$$ $$y=3x+c\tag{2}$$
Now a diagonal of the rhombus would be the angle bisector of the above two lines which would also pass through $(0,y_0)$ and $(1,2)$ so we need to find the equation of the angle bisector. For the equations given below: $$a_{1}x+b_{1}y=c_{1}$$$$a_{2}x+b_{2}y=c_{2}$$ We have the angle bisector formula as:
$$\frac{a_{1}x+b_{1}y-c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{a_{2}x+b_{2}y-c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}$$
First, we get the equation of the angle bisector with $(1)$ and $(2)$ using the above formula and then we use the point $(1,2)$ to find the value of $c$. This will complete the equation and then we can use $(0,y_0)$ to find $y_0$. In case the vertex $A$ lies on the opposite side the point will be $(2,y_0)$.