Answer
See the detailed answer below.
Work Step by Step
First, we need to sketch this problem as we can see below.
We know that the electric field exerted by an infinite line charge at a distance $r$ from it is given by
$$E_{line}=\dfrac{1}{4\pi \epsilon_0}\;\dfrac{2|\lambda|}{r}$$
Now we need to draw the electric field direction as seen below, where $E_x$ is the electric field from the horizontal line and $E_y$ from the vertical line.
Thus, the net electric filed at point $(x,y)$ is given by
$$E_{net}=E_y\;\hat i+E_x\;\hat j $$
$$E_{net}=\dfrac{1}{4\pi \epsilon_0}\;\dfrac{2|\lambda|}{x}\;\hat i+\dfrac{1}{4\pi \epsilon_0}\;\dfrac{2|\lambda|}{y}\;\hat j $$
$$\boxed{E_{net}=\dfrac{2\lambda}{4\pi \epsilon_0}\;\left[\dfrac{1}{x}\;\hat i+ \dfrac{1}{y}\;\hat j\right] }$$
Hence, the electric field strength is given by
$$E=\sqrt{E_x^2+E_y^2}$$
where $E_x=\dfrac{2\lambda}{4\pi \epsilon_0 x}$, and $E_x=\dfrac{2\lambda}{4\pi \epsilon_0 y}$.
$$\boxed{E=\dfrac{2\lambda}{4\pi \epsilon_0 }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}}}$$
