I1.4 In the coordinate plane, the distance from the point $(-C, 0)$ to the straight line $y=x$ is $\sqrt{D}$, find the value of $D$.
The distance from $\mathrm{P}\left(x_{0}, y_{0}\right)$ to the straight line $A x+B y+C=0$ is $\left|\frac{A x_{0}+B y_{0}+C}{\sqrt{A^{2}+B^{2}}}\right|$.
The distance from $(-4,0)$ to $x-y=0$ is $\sqrt{D}=\left|\frac{-4-0+0}{\sqrt{1^{2}+(-1)^{2}}}\right|=\frac{4}{\sqrt{2}}=2 \sqrt{2}=\sqrt{8} ; D=8$
Method 2
$$
\begin{array}{l}
\sqrt{D}=4 \sin 45^{\circ}=\frac{4}{\sqrt{2}} \\
D=\frac{16}{2}=8
\end{array}
$$