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

Given: 

4x− 7y − 3 = 0 … (1) 

2x − 3y + 1 = 0 … (2) 

To find: 

Equation of line passing through the point of intersection of lines. 

Concept Used: 

Point of intersection of two lines. 

Explanation: 

Solving (1) and (2) using cross - multiplication method:

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

⇒ x = - 8 , y = - 5 

Thus, the point of intersection of the given lines is (- 8, - 5). 

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

This line passes through ( - 8, - 5)

\(-\frac{8}{a}-\frac{5}{a}=1\)

⇒ - 8 - 5 = a 

⇒ a = - 13 

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