Question

If the straight line \[\frac{x}{a} + \frac{y}{b} = 1\] passes through the point of intersection of the lines x + y = 3 and 2x − 3y = 1 and is parallel to x − y − 6 = 0, find a and b.

Answer 1

The 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:

\[\frac{x}{- 1 - 9} = \frac{y}{- 6 + 1} = \frac{1}{- 3 - 2}\]

\[ \Rightarrow x = 2, y = 1\]

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).

\[\therefore \frac{2}{a} + \frac{1}{b} = 1\]    ... (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 \[\frac{x}{a} + \frac{y}{b} = 1\]

\[\Rightarrow y = - \frac{b}{a}x + b\]  is equal to the slope of x − y − 6 = 0 or, y = x − 6 

\[\therefore - \frac{b}{a} = 1\]

\[\Rightarrow b = - a\]            ... (4)

From (3) and (4): \[\frac{2}{a} - \frac{1}{a} = 1 \Rightarrow a = 1\]

From (4):
b = −1
∴ a = 1, b = −1