Question

# Find the equation of the line passing through the point of intersection of the lines $4x-7y-3=0$ and $2x-3y+1=0$ that has equal intercepts on the axes.

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Hint: Assume an equation of line having equal intercepts. The intersection point of the other two lines given in the question lies on the assumed line.

Before proceeding with the question, we must know that the equation of a line which is
make intercept of length $'a'$ on $x-$axis and intercept of length $'b'$ on $y-$axis is given by,
$\dfrac{x}{a}+\dfrac{y}{b}=1...........\left( 1 \right)$
It is given in the question that the line has equal intercepts on both the $x$ and $y$ axis. For this line, let
us consider the length of both the intercepts equal to $'a'$. To find the equation of this line, we will
substitute $b=a$ in equation $\left( 1 \right)$. The equation of this line,
\begin{align} & \dfrac{x}{a}+\dfrac{y}{a}=1 \\ & \Rightarrow \dfrac{x+y}{a}=1 \\ & \Rightarrow x+y=a............\left( 2 \right) \\ \end{align}
It is given in the question that this line in equation $\left( 2 \right)$ passes through the intersection
of the lines,
$4x-7y-3=0..........\left( 3 \right)$
And $2x-3y+1=0............\left( 4 \right)$
To find the intersection point, we will solve the two lines with each other. Multiplying line in
equation $\left( 4 \right)$ by $2$, we get,
$4x-6y+2=0...........\left( 5 \right)$
Subtracting line in equation $\left( 3 \right)$ from line in equation $\left( 4 \right)$, we get,
\begin{align} & \left( 4x-6y+2 \right)-\left( 4x-7y-3 \right)=0-0 \\ & \Rightarrow y+5=0 \\ & \Rightarrow y=-5............\left( 6 \right) \\ \end{align}

Substituting $y=-5$ from equation $\left( 6 \right)$ in equation $\left( 3 \right)$, we get,
\begin{align} & 4x-7\left( -5 \right)-3=0 \\ & \Rightarrow 4x+35-3=0 \\ & \Rightarrow 4x=-32 \\ & \Rightarrow x=-8.........\left( 7 \right) \\ \end{align}
From equation $\left( 6 \right)$ and equation $\left( 7 \right)$, we get the intersection points of
lines $4x-7y-3=0$ and $2x-3y+1=0$ $\equiv \left( -8,-5 \right)$.
It is given in the question that the line in equation $\left( 2 \right)$ passes through the intersection
of the lines $4x-7y-3=0$ and $2x-3y+1=0$ i.e. $\left( -8,-5 \right)$. So, substituting $\left( -8,-5 \right)$ in the line in equation $\left( 2 \right)$, we get,
\begin{align} & \left( -8 \right)+\left( -5 \right)=a \\ & \Rightarrow a=-13..........\left( 8 \right) \\ \end{align}
Substituting $a=-13$ from equation $\left( 8 \right)$ in equation $\left( 2 \right)$, we get the
required equation of line,
$x+y=-13$
Or $x+y+13=0$

Note: There is an alternate approach to solve this if one cannot remember the intercept form of the line as in equation $\left( 1 \right)$. The other method from which we can find this equation is, just assume two points $\left( a,0 \right),\left( 0,b \right)$ and apply a two-point form formula to get the equation of line.