Q. Find the equation of line passing through the pt of intersection of the lines- $3x-y=5$ and $x+3y=1$ and making equal and positive intercepts on both the axes.
A. To find the pt of intersection substitute $x=1-3y$ in $3x-y=5$. Thus, $y= \frac{-1}{5}$ and $x= \frac{8}{5}$.
Thus, $b(x-h)-a(y-k)=0$ where a and b are the x and y intercepts respectively. And h and k the pt of intersection the x and y axis respectively.
Since, $a=b=z(say)$- $$z[x-h-y+k]=0$$ $$x- \frac{8}{5} -y - \frac{1}{5}=0$$ $$5x-5y-7=0$$
But the answer key says it to be $5x+5y-7=0$
Also, I would like the derivation of the formula $b(x-h)-a(y-k)=0$ which was simply stated in my book.
Hint:
A line passing thorough the point $P=(\frac{8}{5},\frac{-1}{5})$ (and not parallel to the $y$ axis) has equation: $$ y+\frac{1}{5}=m\left(x-\frac{8}{5}\right) $$ and, if this line has equals positive intercepts with the $x$ and $y$ axis, than $m=-1$, so it has equation: $$ y+\frac{1}{5}+\left(x-\frac{8}{5}\right)=0 $$