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प्रश्न
Find the equation of the straight line passing through the point (2, 1) and bisecting the portion of the straight line 3x − 5y = 15 lying between the axes.
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उत्तर
The equation of the line in intercept form is \[\frac{x}{a} + \frac{y}{b} = 1\].
The line passes through (2, 1).
∴ \[\frac{2}{a} + \frac{1}{b} = 1\] ... (1)
Let the line 3x − 5y = 15 intersect the x-axis and the y-axis at A and B, respectively.
At x = 0 we have,
0 − 5y = 15
\[\Rightarrow\] y = −3
At y = 0, we have,
3x − 0 = 15
\[\Rightarrow\] x = 5
\[\therefore A \equiv \left( 0, - 3 \right) \text { and } B \equiv \left( 5, 0 \right)\]
The midpoint of AB is \[\left( \frac{5}{2}, - \frac{3}{2} \right)\].
Clearly, the point
\[\left( \frac{5}{2}, - \frac{3}{2} \right)\] lies on the line \[\frac{x}{a} + \frac{y}{b} = 1\].
∴ \[\frac{5}{2a} - \frac{3}{2b} = 1\] ... (2)
Using \[\frac{3}{2} \times eq (1) + eq (2)\] we get,
\[\frac{3}{a} + \frac{5}{2a} = \frac{3}{2} + 1\]
\[ \Rightarrow a = \frac{11}{5}\]
For a = \[\frac{11}{5}\] we have,
\[\frac{10}{11} + \frac{1}{b} = 1\]
\[ \Rightarrow b = 11\]
Hence, the equation of the required line is
\[\frac{5x}{11} + \frac{y}{11} = 1\]
\[ \Rightarrow 5x + y = 11\]
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