I am new to coordinate geometry. I am not able to solve this question, as I have no idea, from where to start.
Question:
Two sides of a 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) $ and the vertex $ A $ is on the $ y $ - axis , find possible coordinates of $ A $ .
On
(see figures below representing the two solutions).
Let the vertices $ABCD$ be taken in the direct (trigonometric) orientation. Let $I$ denote the rhombus' center.
The equations of the lines with the given slopes passing by $A(0,a)$ are :
$$y-7x-a=0 \ \ \text{and} \ \ y-x-a=0$$
Let us recall that the distance of a point $P(x_0,y_0)$ to a straight line L with equation $ux+vy+w=0$ is
$$\tag{1}\text{dist(P,L)=}\dfrac{|ux_0+vy_0+w|}{\sqrt{u^2+v^2}}$$
A necessary and sufficient condition of "rhombusity" (!) is that diagonal $AI$ is the angle bissector of angle between $AB$ and $AD$, i.e.,
$$\text{dist(I,AB)=dist(I,AD)}$$
Thus, using (1), we need to solve the following equation:
$$\dfrac{|y_0-7x_0-a|}{\sqrt{1^2+7^2}}=\dfrac{|y_0-x_0-a|}{\sqrt{1^2+1^2}} \ \text{with} \ (x_0,y_0)=(1,2)$$
This gives $|-5-a|=5|1-a|$, out of which one gets two solutions
$$a=0 \ \ \ \text{and} \ \ \ a=2.5$$
(Figures representing the two cases $a=0$ and $a=2.5$)
Let $A(0,a)$ and $E(1,2)$.
Thus, $m_{AE}=2-a$ and we obtain: $$\left|\frac{2-a-1}{1+(2-a)\cdot1}\right|=\left|\frac{2-a-7}{1+(2-a)\cdot7}\right|.$$ I hope you could end it now.