$\begin {array} {1 1} (A)\;13x-13y=6 & \quad (B)\;13x+13y=6 \\ (C)\;6x+6y=13 & \quad (D)\;6x-6y=13 \end {array}$

- Equation of the line having intercepts $a$ and $b$ on the axes is $ \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1$

Let the equation of the line having equal intercepts on the axes is

$ \large\frac{x}{a}$$+ \large\frac{y}{a}$$=1$

(i.e.,) $x+y=a$----------(1)

Given equation of the line is

$ 4x+7y-3=0\: and \: 2x-3y+1=0$

Let us solve the above equations for $x$ and $y$.

$ \qquad 4x+7y=3$-------(2)

$ ( \times 2) 2x-3y=-1$--------(3)

$ \qquad 4x+7y=3$

$ \qquad 4x-6y=-2$

$ \qquad \qquad 13y=5 \Rightarrow y = \large\frac{5}{13}$

$4x+7 \bigg( \large\frac{5}{13} \bigg)$$ = 3$

$ \Rightarrow 4x=3-\large\frac{35}{13}$

$ \Rightarrow x = \large\frac{1}{13}$

Hence the point of intersection is $ \bigg( \large\frac{1}{13}$$, \large\frac{5}{13} \bigg)$

Since equation (1) passes through the point $ \bigg( \large\frac{1}{13}$$, \large\frac{5}{13} \bigg)$.

Substituting these values for $x$ and $y$ in equation (1) we get,

$ \large\frac{1}{13}$$+ \large\frac{5}{13}$$=a$

$ \Rightarrow \large\frac{6}{13}$$=a$ Substituting the value of $ 'a'$ in equation (1)

$x+y=\large\frac{6}{13}$

(i.e.,) $13x+13y=6$

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