How does one find the equation of all lines passing through a point (Ex. $(6, -1)$), satisfying the condition that the product of their $x$ and $y$ intercepts must equal some number $c$ (Ex. $3$)?
As far as I understand this can be conceptualized as finding the equation of the line containing the points $(6,-1), (a,0), (0,b)$ where $a$ is the $x$-intercept, $b$ is the $y$-intercept and $ab=3$.
I've tried finding the slope with $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ and have gotten $m=\frac{-1}{6-a}$ and $m=\frac{-1-b}{6}$. I've also solved for $a$ and $b$ in terms of $m$ and tried susbstituting these values into the equation of a line ($y=mx+b$) but I just cant eliminate enough variables to solve for anything useful. Feel like I'm missing something obvious, I'm not even sure if there is more than one equation that satisfies the conditions.
Best Answer
Intercept equation of a line:
$x/a+y/b=1$;
This line passes through $(x_0,y_0):$
$x_0/a+y_0/b=1$;
Given: $c=ab$;
Then
$x_0/a+(ay_0/c)=1$;
Solve for $a$:
$y_0 a^2-ac+cx_0=0$;
Quadratic in $a$: $a_{1,2}$:
Lines:
$x/a_{1,2} +y/(c/a_{1,2})=1.$