The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th

The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th
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(c) 25x + 25y – 4 = 0 The point of intersection of the given lines can be obtained by solving the equations of the two lines simultaneously. 100x + 50y = 1                    ...(i) 75x + 25y = –3                  ...(ii) Eqn (i) – 2 × Eqn (ii) ⇒ (100x + 50y) – (150x + 50y) = 1 – (–6) ⇒ – 50x = 7 ⇒ x = \(rac{-7}{50}\)100 x \(rac{-7}{50}\) + 50y  = 1 ⇒ 50y = 1 + 14 = 15 ⇒ y = \(rac{15}{50}\) = \(rac{3}{10}\)Let the required line make x-intercept = y-intercept = a. Then eqn of required line is \(rac{x}{a}\) + \(rac{y}{a}\) = 1 ⇒ x + y = aSince it passes through \(\bigg(\)\(rac{-7}{50}\), \(rac{3}{10}\)\(\bigg)\), therefore⇒ \(rac{-7}{50}\) + \(rac{3}{10}\) = a ⇒ \(rac{-7+15}{50}\) = a ⇒ a = \(rac{8}{50}\) = \(rac{4}{25}\)∴ Eqn of required line: x + y = \(rac{4}{25}\) ⇒ 25 + 25y – 4 = 0.
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