Consider the equation of lines, 3x−4y+1=0 And, 5x+y−1=0 Thus, 3x−4y+1+λ(5x+y−1)=0 (3+5λ)x+(−4+λ)y+1−λ=0 The given lines have equal intercepts. 3+5λ=−4+λ λ=−
7
4
Therefore, the equation of line is, 23x+23y=11
x
11
23
+
y
11
23
=1 The area of the triangle formed by the line and the coordinate axes is, A=