What we have is a line with positive slope, passing through the point $(-2, 5)$, with the length of the segment on this line from the point where it intersects the negative $x$-axis at $(x_0, 0)$ and the positive $y$-axis at $(0, y_0)$ equal to $7\sqrt 2$.
We know then, that $$x_0^2 + y_0^2 = (7\sqrt 2)^2 = 98\tag{1}$$
We also know that the distance from $(x_0, 0)$ to $(-2, 5)$ plus the distance from $(-2, 5)$ to the point $(0, y_0)$ is equal to $7\sqrt 2$: $$\sqrt {25 + (x_0+2)^2} + \sqrt{4+(y_0-5)^2} = 7\sqrt 2\tag{2}.$$
Solving $(1)$ and $(2)$ simultaneously gives us the real solutions $x_0 = -7$, and $y_0 = 7$. With these values, we must have that the slope of each of these "segments" comprising the entire length are equal: $$m = \frac{5-0}{-2 - x_0} = \frac{5-y_0}{-2-0}= 1$$
Now, given your point on the line $(-2, 5)$ and the slope of $m = 1$, we can construct the equation of the line: $$y - 5 = x + 2 \iff y = x+7$$
Expanding upon my comment, and following the same basic strategy as in this answer, we can get the equation mechanically using the determinant for the five-point conic through $P=(P_x,P_y)$, $Q=(Q_x,Q_y)$, $R=(R_x,R_y)$, $S=(S_x,S_y)$, $T=(T_x,T_y)$:
$$\left|\begin{array}{c,c,c,c,c,c}
x^2 & y^2 & x y & x & y & 1 \\
P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \\
Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \\
R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \\
S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \\
T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \\
\end{array}\right| = 0 \tag{$\star$}$$
It happens to be convenient to have the origin at the midpoint of the given points, so let us take
$$P = (d\cos\theta,d\sin\theta) \qquad Q = (-d\cos\theta, -d\sin\theta) \tag{1}$$
Two more points come from infinitesimally-displacing $P$ and $Q$ along their tangent lines (in directions, say, $\phi$ and $\psi$, respectively):
$$R = P + (p\cos\phi, p\sin\phi) \qquad S = Q + (q\cos\psi,q\sin\psi) \tag{2}$$
for "very small" $p$ and $q$. The fifth point is provided by the specific geometry of the parabola; if $M$ is the midpoint of $P$ and $Q$, and $N$ is the point where the tangent lines at $P$ and $Q$ meet, then the midpoint of $M$ and $N$ lies on the parabola. Thus, we can take
$$T = \frac{d}{2\sin(\phi-\psi)}\left(\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta), \sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta)\right) \tag{3}$$
Substituting into $(\star)$, and letting a computer algebra system crunch some symbols, we factor-out and cancel a $p$ and $q$, then set the remaining $p$s and $q$s to zero. Canceling factors of $d^2 \csc^2(\phi-\psi) \sin(\phi-\theta)\sin(\psi - \theta)$, we have
$$\begin{align}
0 &= \phantom{2}x^2 (\sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta))^2 \\
&+ \phantom{2}y^2 (\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta))^2 \\
&-2x y (\sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta))(\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta)) \\
&-4 x d \sin\theta \sin(\phi-\psi) \sin(\phi-\theta) \sin(\psi-\theta) \\
&+4 y d \cos\theta \sin(\phi-\psi) \sin(\phi-\theta) \sin(\psi-\theta) \\
&-4 d^2 \sin^2(\phi-\theta)\sin^2(\psi-\theta)
\end{align} \tag{$\star\star$}$$
Best Answer
I am assuming that you mean the y-cut is $0$. In other words, the equation of the line is of the form $$y=mx$$ where there is no constant term. I will also only consider lines and not line segments, as it is obvious that a line segment could be defined in such a way that it only falls in one quadrant (See Steven Stadnicki's comment). Then, I agree with your list of properties, except for the following amendments:
If $0<m<1$, the line lies at an angle of between $0 ^{\circ}$ and $45^{\circ}$ or between $180^{\circ}$ to $225^{\circ}$ to the horizontal.
If $m>1$, the line lies at an angle of between $45 ^{\circ}$ and $90^{\circ}$ or between $225^{\circ}$ and $270^{\circ}$ to the horizontal.
If $m<0$, the line lies at an angle of between $90 ^{\circ}$ and $180^{\circ}$ or between $270^{\circ}$ and $360^{\circ}$ to the horizontal.