Given:
The lines x + y = 4 and 2x – 3y = 1
The equation of the straight line passing through the point of intersection of x + y = 4 and 2x − 3y = 1 is
x + y − 4 + λ(2x − 3y − 1) = 0
(1 + 2λ)x + (1 − 3λ)y − 4 − λ = 0 … (1)
y = – [(1 + 2λ) / (1 – 3λ)]x + [(4 + λ) / (1 – 3λ)]
The equation of the line with intercepts 5 and 6 on the axis is
x/5 + y/6 = 1 …. (2)
So, the slope of this line is -6/5
The lines (1) and (2) are perpendicular.
∴ -6/5 × [(-1 + 2λ) / (1 – 3λ)] = -1
λ = 11/3
Now, substitute the values of λ in (1), we get the equation of the required line.
(1 + 2(11/3))x + (1 – 3(11/3))y − 4 – 11/3 = 0
(1 + 22/3)x + (1 – 11)y – 4 – 11/3 = 0
25x – 30y – 23 = 0
∴ The required equation is 25x – 30y – 23 = 0
