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

We have,
4x − 7y − 3 = 0     ... (1)
2x − 3y + 1 = 0     ... (2)
Solving (1) and (2) using cross-multiplication method:

\[\frac{x}{- 7 - 9} = \frac{y}{- 6 - 4} = \frac{1}{- 12 + 14}\]

\[ \Rightarrow x = - 8, y = - 5\]

Thus, the point of intersection of the given lines is \[\left( - 8, - 5 \right)\].

Now, the equation of a line having equal intercept as a is \[\frac{x}{a} + \frac{y}{a} = 1\] .

This line passes through \[\left( - 8, - 5 \right)\].

\[\therefore \frac{- 8}{a} - \frac{5}{a} = 1\]

\[ \Rightarrow - 8 - 5 = a\]

\[ \Rightarrow a = - 13\]

Hence, the equation of the required line is \[\frac{x}{- 13} + \frac{y}{- 13} = 1 \text { or } x + y + 13 = 0 .\]