Given:
lines are x + y = 3 and 2x − 3y = 1.
To find: a and b.
Concept Used:
Point of intersection of two lines.
Explanation:
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 \(\frac{x}{a}+\frac{y}{b}=1\) passes through (2, 1).
∴ \(\frac{2}{a}+\frac{1}{b}=1\) ....(3)
It is also given that the line \(\frac{x}{a}+\frac{y}{b}=1\) is parallel to the line x − y − 6 = 0.
Hence, Slope of \(\frac{x}{a}+\frac{y}{b}=1\) ⇒ y = \(-\frac{b}{a}x+b\) is equal to the slope of x − y − 6 = 0 or, y = x – 6
∴ - b/a = 1
⇒ b = - a … (4)
From (3) and (4):
∴ \(\frac{2}{a}+\frac{1}{b}=1\)
⇒ a = 1
From (4):
b = −1
∴ a = 1,
b = −1
Hence, a = 1, b = - 1
