The line L given by x/5 + y/b = 1 passes through the point (13, 32) . The line K is parallel to L and has the equation x/c + y/3 = 1 . Then the distan - Mathematics MCQ AIEEE 2026 | ExamTest.live

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The line L given by \frac{x}{5} + \frac{y}{b} = 1 passes through the point (13, 32) . The line K is parallel to L and has the equation \frac{x}{c} + \frac{y}{3} = 1 . Then the distance between L and K is:

Solution & Step-by-step Explanation

1. L passes through (13, 32) : \frac{13}{5} + \frac{32}{b} = 1 ⟹ \frac{32}{b} = 1 - \frac{13}{5} = - \frac{8}{5} . b = \frac{32 \times 5}{- 8} = - 20 .Equation of L : \frac{x}{5} - \frac{y}{20} = 1 ⟹ 4 x - y - 20 = 0 .2. K is parallel to L : Slope of K = - 3/ c . Slope of L = 4 . - 3/ c = 4 ⟹ c = - 3/4 .Equation of K : \frac{x}{- 3/4} + \frac{y}{3} = 1 ⟹ - \frac{4 x}{3} + \frac{y}{3} = 1 ⟹ - 4 x + y = 3 ⟹ 4 x - y + 3 = 0 .3. Distance d = \frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}} = \frac{∣ - 20 - 3∣}{4 ^{2} + ( - 1 ) ^{2}} = \frac{23}{17} .

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The line L given by \frac{x}{5} + \frac{y}{b} = 1 passes through the point (13, 32) . The line K is parallel to L and has the equation \frac{x}{c} + \frac{y}{3} = 1 . Then the distance between L and K is:

17

17/15

23/ 17

23/15

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