Given lines are x + y = 3 and 2x − 3y = 1.
x + y − 3 = 0 … (1)
2x − 3y − 1 = 0 … (2)
Solving (1) and (2) using cross - multiplication method,
\(\frac x{-1-9} = \frac y{-6 +1}= \frac 1{-3-2}\)
⇒ x = 2 , y = 1
Thus, the point of intersection of the given lines is (2, 1).
It is given that the line \(\frac{x}a + \frac yb =1\) passes through (2, 1).
∴ \(\frac 2a + \frac 1b = 1\) ....(3)
It is also given that the line \(\frac{x}a + \frac yb =1\) is parallel to the line x − y − 6 = 0.
Hence, Slope of \(\frac{x}a + \frac yb =1\) ⇒ y = \(- \frac bax + b\) is equal to the slope of x − y − 6 = 0 or, y = x – 6
∴ \(-\frac ba\) = 1
⇒ b = - a … (4)
From (3) and (4),
∴ \(\frac 2a + \frac 1b = 1\)
⇒ a = 1
From (4),
b = −1
∴ a = 1,
b = −1
Hence, a = 1, b = - 1.
