Solving lines L1(3x+2y=14) and L2(5x−y=6).
⇒x=314−2y=56+y
⇒70−10y=18+3y
⇒52=13y
⇒y=4
⇒x=314−8=2
⇒A≡(2,4)
Now, solving lines L3(4x+3y=8) and L4(6x+y=5)
⇒x=48−3y=65−y
⇒48−18y=20−4y
⇒28=14y
⇒y=2
⇒x=48−6=21
⇒B≡(21,2)
So, equation of AB is given by,
AB:(y−4)=(21−22−4)(x−2)
⇒AB:(y−4)=(34)(x−2)
⇒AB:(y−4)=(34)(x−2)
⇒3y−12=4x−8
⇒4x−3y+4=0
The distance of P(5,−2) from the line AB is given by,
D=∣54(5)−3(−2)+4∣
⇒D=6 units.