Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0

Consider the equation of the line having equal intercepts on the axes as

\( \dfrac{x}{a}+\dfrac{y}{a}=1\)

It can be written as

x + y = a ….. (1)

By solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 we get

x = \( \dfrac{1}{13}\) and y = \( \dfrac{5}{13}\)

(\( \dfrac{1}{13}\), \( \dfrac{5}{13}\)) is the point of intersection of two given lines

We know that equation (1) passes through point (\( \dfrac{1}{13}\),\( \dfrac{5}{13}\))

\( \dfrac{1}{13}\) +\( \dfrac{5}{13}\) = a

a =\( \dfrac{6}{13}\)

So the equation (1) passes through (\( \dfrac{1}{13}\), \( \dfrac{5}{13}\))

\( \dfrac{1}{13}\) +\( \dfrac{5}{13}\) = a

We get

a = \( \dfrac{6}{13}\)

Here the equation (1) becomes

x + y = \( \dfrac{6}{13}\)

13x + 13y = 6

Hence, the required equation of the line is 13x + 13y = 6.