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प्रश्न
Find the equation of straight line passing through (−2, −7) and having an intercept of length 3 between the straight lines 4x + 3y = 12 and 4x + 3y = 3.
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उत्तर
Here,
\[\left( x_1 , y_1 \right) = A\left( - 2, - 7 \right)\]
So, the equation of the line is
\[\frac{x - x_1}{cos\theta} = \frac{y - y_1}{sin\theta}\]
\[ \Rightarrow \frac{x + 2}{cos\theta} = \frac{y + 7}{sin\theta}\]
Let the required line intersect the lines 4x + 3y = 3 and 4x + 3y = 12 at P1 and P2.
Let AP1 = r1 and AP2 = r2
Then, the coordinates of P1 and P2 are given by
\[\frac{x - x_1}{cos\theta} = \frac{y - y_1}{sin\theta}={r_1}\] and \[ \Rightarrow \frac{x + 2}{cos\theta} = \frac{y + 7}{sin\theta}= {r_2}\], respectively.
Thus, the coordinates of P1 and P2 are \[\left( - 2 + r_1 cos\theta, - 7 + r_1 sin\theta \right) \text { and } \left( - 2 + r_2 cos\theta, - 7 + r_2 sin\theta \right)\], respectively.
Clearly, the points P1 and P2 lie on the lines 4x + 3y = 3 and 4x + 3y = 12
\[4\left( - 2 + r_1 cos\theta \right) + 3\left( - 7 + r_1 sin\theta \right) = 3 and 4\left( - 2 + r_2 cos\theta \right) + 3\left( - 7 + r_2 sin\theta \right) = 12\]
\[ \Rightarrow r_1 = \frac{32}{4cos\theta + 3sin\theta} \text { and } r_2 = \frac{41}{4cos\theta + 3sin\theta}\]
\[\text { Here }, A P_2 - A P_1 = 3 \Rightarrow r_2 - r_1 = 3\]
\[ \Rightarrow \frac{41}{4cos\theta + 3sin\theta} - \frac{32}{4cos\theta + 3sin\theta} = 3\]
\[ \Rightarrow 3 = 4cos\theta + 3sin\theta\]
\[ \Rightarrow 3\left( 1 - sin\theta \right) = 4cos\theta\]
\[ \Rightarrow 9\left( 1 + \sin^2 \theta - 2sin\theta \right) = 16 \cos^2 \theta = 16\left( 1 - \sin^2 \theta \right)\]
\[ \Rightarrow 25 \sin^2 \theta - 18sin\theta - 7 = 0\]
\[ \Rightarrow \left( sin\theta - 1 \right)\left( 25sin\theta + 7 \right) = 0\]
\[ \Rightarrow sin\theta = 1, sin\theta = - \frac{7}{25}\]
\[ \Rightarrow cos\theta = 0, cos\theta = \frac{24}{25}\]
Thus, the equation of the required line is
\[x + 2 = 0\text { or } \frac{x + 2}{\frac{24}{25}} = \frac{y + 7}{\frac{- 7}{25}}\]
\[ \Rightarrow x + 2 = 0 \text { or } 7x + 24y + 182 = 0\]
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