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.

Answer 1

Let the equation of the line having equal intercepts on the axes be  \(\frac{x}{a} +\frac{ y}{a} = 1\)
\(x + y = a ….. (1)\)
On solving equations \(4x + 7y - 3 = 0\) and \(2x - 3y + 1 = 0\), we obtain

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

\(∴ (\frac{1}{13}, \frac{5}{13})\)  is the point of intersection of two given lines.
Since equation (1) passes through point \((\frac{1}{13},\frac{ 5}{13})\)
\(\frac{1}{13} +\frac{ 5}{13} = a\)

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

∴ Equation (1) becomes 
\(x + y =\frac{ 6}{13}\)
\(13x + 13y = 6\)

Thus, the required equation of the line is \(13x + 13y = 6.\)