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:

$$\begin{gathered} \frac{x}{-7-9}=\frac{y}{-6-4}=\frac{1}{-12+14} \\ \Rightarrow x=-8,y=-5 \end{gathered}$$

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 $\left(-8,-5\right)$.

$$\begin{gathered} \therefore \frac{-8}{a}-\frac{5}{a}=1 \\ \Rightarrow -8-5=a \\ \Rightarrow a=-13 \end{gathered}$$

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