Find all $k$ when lines cross each other inside quadrant 1

Question

\begin{align} kx+5y &= -13\\ x+ky &= 1 \end{align} Find all $k$ when these lines cross each other inside quadrant 1.

I used Cramer's rule. I got $x = \dfrac{-13k-5}{k^2-5}$ and $y = \dfrac{k+13}{k^2-5}$. After that I got stuck.

Answer 1

I agree that parametric linear system of two equations with two unknowns: $$\cases{kx+5y=-13\\x+ky=1}$$ has solutions

$$\tag{1}x=\dfrac{-13k-5}{k^2-5}, \ \ y=\dfrac{k+13}{k^2-5}.$$

(Cramer's rule with principal determinant $\begin{vmatrix}k&5\\1&k\end{vmatrix}=k^2-5$, etc.).

Conditions on parameter $k$ for $(x,y)$ to be in the first quadrant are ruled by an array of signs, with limits governed by the sign changes of the different numerators and denominators in (1).

$$\begin{array}{c|ccccccc}k&&-13&&-\sqrt{5}&&-5/13&&+\sqrt{5} &&\\ \hline k^2-5&+&&+&&-&&- &&+\\-13k-5&+&&+&&+&&- &&-\\k+13&?&&?&&?&&? &&?\\x&+&&+&&-&&+ &&-\\y&?&&?&&?&&? &&?\end{array}$$

(I leave you some work to do ...).

Final step: examine the different intervals $(-\infty,-13)$, etc. The solution gathers those for which, simultaneously $x>0$ and $y>0$.

Remarks : for a complete understanding of the issue, one can observe

1) that the first (resp. second) equation account for a pencil of lines through point $B$ (resp. point $A$) as can be seen on the figure below.

2) that the intersection points given by (1) are such that

$$x+13y-164=0$$

which means they are aligned.