Let a variable line of slope $ m0 $ passing through the point $ (4, -9) $ intersect the coordinate axes at the points $ A $ and $ B $. The minimum value of the sum of the distances of $ A $ and $ B $ from the origin is:

Question

Let a variable line of slope $ m>0 $ passing through the point $ (4, -9) $ intersect the coordinate axes at the points $ A $ and $ B $. The minimum value of the sum of the distances of $ A $ and $ B $ from the origin is:

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The equation of the line is: $$ y + 9 = m(x - 4). $$ Find $ A $ and $ B $ (intersection points with the axes): At $ y = 0 $, $ x = \frac{9}{m} + 4 \implies A\left(\frac{9}{m} + 4, 0\right) $. At $ x = 0 $, $ y = -9 - 4m \implies B(0, -9 - 4m) $. The sum of distances: $$ OA + OB = \sqrt{\left(\frac{9}{m} + 4\right)^2} + \sqrt{(-9 - 4m)^2}. $$ Using AM-GM inequality, the minimum value occurs when: $$ m = \frac{3}{2}. $$ Substitute $ m $ to get: $$ OA + OB = 25. $$

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Let a variable line of slope \( m>0 \) passing through the point \( (4, -9) \) intersect the coordinate axes at the points \( A \) and \( B \). The minimum value of the sum of the distances of \( A \) and \( B \) from the origin is:

The Correct Option is A

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