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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.

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.