4. Draw a line through point $P(-1,3)$, so that the triangle formed with the two coordinate axes has an area of 5. How many such lines can be drawn?
(A) 4
(B) 3
(C) 2
(D) 1
4.C.

Let the line $l: y=kx+b$. Since $P(-1, 3)$ lies on the line $l$, we have $3=-k+b$, hence $b=k+3$.
Therefore, $l$ is $y=kx+k+3$.
The intersection points of $l$ with the two coordinate axes are $A\left(-\frac{k+3}{k}, 0\right)$, $B(0, k+3)$, so the area of the triangle formed by $l$ and the two coordinate axes is
$$
S_{\triangle AOB}=\frac{1}{2}|OA| \cdot|OB|=\frac{1}{2}\left|-\frac{k+3}{k}\right| \cdot|k+3|=5 \text{, }
$$

which means $(k+3)^{2}=10|k|$.
When $k>0$, equation (1) becomes $k^{2}-4k+9=0$, which has no real solutions;

When $k<0$, equation (1) becomes $k^{2}+16k+9=0$, which has two real solutions $k_{1,2}=-8 \pm \sqrt{55}$.

Therefore, there are two lines passing through the point $P(-1,3)$ and forming a triangle with the coordinate axes with an area of 5.