equation of the line that passes through the point of intersection of the lines $3x-4y+1=0$ and $5x+y-1=0$ and cuts same length from both the axes.

Question

Find the equation of the line that passes through the point of intersection of the lines $3x-4y+1=0$ and $5x+y-1=0$ and cuts the same length from both the axes.
$\bf{Try:} $ Since the line cuts the same length from both the axes, let the equation of the line be $\frac xa +\frac ya=1$. The given two lines intersect at the point $\left(\frac 3{23},\frac 8{23}\right)$. Putting this point in the above equation we get $a =\frac {11} {23}$. Hence the equation of the line becomes $23x +23y=11$. But it has two given answers $23x +23y=11$ and $23x-23y+5=0$. I'm unable to find how the second equation comes and why. Please give reason and describe. Thanks in advance.

Answer 1

Consider the two lines - \begin{gather*} L_{1} :\frac{x}{a} +\frac{y}{a} =1\\ L_{2} :\frac{x}{a} -\frac{y}{a} =1 \end{gather*} L1 intersects the axes at $(a,0)$ and $(0,a)$. What is the distance between these two points, using the distance formula? This is the intercept of L1.

L2 intersects the axes at $(a,0)$ and $(0,-a)$. What is the distance between these two points? This is the intercept of L2.

Notice anything strange?

Both the intercepts turn out to be equal, by the distance formula!

Using this fact, can you redo your problem so you get both the correct answers?

Answer 2

The required line is $$(3x-4y+1)+t(5x+y-1)=0 \implies (3+5t)x+(-4+t)y+(1-t)=0~~~~(1)$$ Its cut on $x$-axis is $A=\frac{t-1}{3+5t}$ and on $y$ axis it is $B=\frac{t-1}{t-4}.$ $A=B \implies t=-7/4$, we get the Eq. of line from (1) as $$x+y=11/23$$

EDIT: $|A|=|B|$ the other equation shuld be $A=-B \implies 3+5t=-(t-4) \implies t=1/6$ this choice gives $$x-y=-5/23$$