# 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.

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

Toolbox:
• 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.,) $13x+13y=6$