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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.
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Solution
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
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