$d(x,y) = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2 }.$ $(\mathbb R^2,d)$ is a metric space . I am facing problem while solving its triangular inequality. In case of collinear points equality holds but what happen if points are not collinear.

For three point $x,y,z \in \mathbb{R}^2$, you need to show that
$$d(x,y) \le d(x,z)+d(y,z)$$
Applying your metric, gives us 
$$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \le \sqrt{(x_1-x_2)^2+(z_1-z_2)^2} + \sqrt{(y_1-y_2)^2+(z_1-z_2)^2}$$
Let $\Delta x := x_1-x_2$, $\Delta y := y_1-y_2$ and $\Delta z := z_1-z_2$. We need to prove that
$$\sqrt{(\Delta x)^2+(\Delta y)^2} \le \sqrt{(\Delta x)^2+(\Delta z)^2} + \sqrt{(\Delta y)^2+(\Delta z)^2}$$
Since both the left-hand and the right-hand sides are non-negative, we can square both sides without worrying about the inequality: $a \le b \iff a^2 \le b^2$ for all $a,b \ge 0$. This gives
$$(\Delta x)^2+(\Delta y)^2 \le (\Delta x)^2+(\Delta z)^2 + 2\sqrt{(\Delta x)^2+(\Delta z)^2}\sqrt{(\Delta y)^2+(\Delta z)^2} + (\Delta y)^2+(\Delta z)^2$$
$$0 \le 2(\Delta z)^2 + 2\sqrt{(\Delta x)^2+(\Delta z)^2}\sqrt{(\Delta y)^2+(\Delta z)^2}$$
It should be clear that this final inequality is true for all $\Delta x$, $\Delta y$ and $\Delta z$.
