Given lines are 4x + 1y-3 = 0 …………….(1) 2x – 3y + 1 = 0 ………………. (2) First, find the point of intersection of (1) and (2), ∴ 3 x (1) + 7 x (2) gives 12 x + 14x – 9 + 7 = 0
⇒ 26x - 2 = 0 ⇒ x = 1/13
Put x = 1/13 in (2), we get
2/13 - 3y + 1 = 0 ⇒ 3y = 15/13 ⇒ y = 5/13
∴ (1) and (2) meet at (1/13, 5/13)
Since the required line has equal intercepts on the axes then its equation is of the form,
x/a + y/a = 1 (∵a = b)
i.e., x + y = a ...(3)
the line (3) passes through (1/13, 5/13) then
1/13 + 5/13 = a ⇒ a = 6/13
Putting the value of 'a' in(3), we get
x + y = 6/13
i.e., 13x + 13y = 6