Compute conditional probability $P(x\geq \frac 12 \mid y=2x)$ 
Consider the PDF $$f(x,y)=\frac 12 xy\quad\text{if}\quad 0\leq x\leq y\leq2,\qquad f(x,y)=0\quad \text{otherwise}$$
  How to compute
  $$P\left(\left.x\geq \frac 12 \right| y=2x\right)\ ?$$

I can solve $P(x \in A | y \in B )$  when $A$ and $B$ are intervals
but this conditioning uses $y=2x$, which is a line. 
How to control $x$ and $y$ ?
 A: Here are two natural interpretations of the question, which lead to different solutions and show the question is not well founded.

First, $Y=2X$ means that $Z=2$ with $Z=Y/X$ hence this could be asking for $u(2)$, where $$P(X\geqslant \tfrac12\mid Z)=u(Z)$$
To compute this, note that the joint PDF of $(X,Z)$ is $$f_{X,Z}(x,z)=\tfrac12x^3z\quad\text{on}\quad x>0,\ z>1,\ x<\tfrac2z$$
hence, for every $z>1$, the conditional PDF is $$f_{X\mid Z}(x\mid z)\propto  x^3\quad\text{on}\quad 0<x<\tfrac2z$$ that is,
$$f_{X\mid Z}(x\mid z)=\tfrac14z^4x^3\quad\text{on}\quad 0<x<\tfrac2z$$
In particular, $$f_{X\mid Z}(x\mid 2)=4x^3\quad\text{on}\quad 0<x<1$$ hence 
$$P(X\geqslant \tfrac12\mid Z=2)=\int_{1/2}^14x^3dx=1-\tfrac1{16}$$

But $Y=2X$ also means that $T=0$ with $T=Y-2X$ hence this could as well be asking for $v(0)$, where
$$P(X\geqslant \tfrac12\mid T)=v(T)$$
To compute this, note that the joint PDF of $(X,T)$ is $$f_{X,T}(x,t)=\tfrac12x(t+2x)\quad\text{on}\quad 0<x<t+2x<2$$
hence, for every $|t|<2$, the conditional PDF is $$f_{X\mid T}(x\mid t)\propto  x(t+2x)\quad\text{on}\quad x>0,\ x>-t,\ x<1-t/2$$
In particular, $$f_{X\mid T}(x\mid 0)=3x^2\quad\text{on}\quad 0<x<1$$
hence $$P(X\geqslant \tfrac12\mid T=0)=\int_{1/2}^13x^2dx=1-\tfrac18$$

Hmmm... $1-\tfrac1{16}\ne1-\tfrac1{8}$, right? So, which solution is correct? Both are, and perhaps yet a third one...
