A line through the point A(2,4) intersects the line `x+y=9` at the point. The minimum distance of AP , isA. `(5)/(sqrt(2))`B. `(7)/(sqrt(2))`C. `(2)/(sqrt(2))`D. `(1)/(sqrt(2))`

A line through the point A(2,4) intersects the line `x+y=9` at the poi

1.	A line through the point A(2,4) intersects the line `x+y=9` at the point. The minimum distance of AP , isA. `(5)/(sqrt(2))`B. `(7)/(sqrt(2))`C. `(2)/(sqrt(2))`D. `(1)/(sqrt(2))`
Answer» Correct Answer - C The equation of a line passing through A(2,4) is `(x-2)/(cos theta)=(y-4)/(sin theta)` Suppose it cuts the line `x+y=9` at point P whose coordinates are given by `(x-2)/(cos theta)=(y-4)/(sin theta)=r` I.e. `x=2+r cos theta, y=4+r sin theta` `therefore 2+rcos theta+4+rsin theta=9` `implies r(cos theta+sin theta)=3` `implies r=(3)/(cos theta+sin theta)` `implies r ge (3)/(sqrt(2)) [because cos theta + sin theta le sqrt(2) therefore (1)/(cos theta + sin theta ge (1)/(sqrt(2)))]` Hence , the minimum value of AP is `(3)/(sqrt(2))`

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A line through the point A(2,4) intersects the line `x+y=9` at the point. The minimum distance of AP , isA. `(5)/(sqrt(2))`B. `(7)/(sqrt(2))`C. `(2)/(sqrt(2))`D. `(1)/(sqrt(2))`

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