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प्रश्न
Find the point on the straight line 2x+3y = 6, which is closest to the origin.
उत्तर
The equation of line is given as 2x+3y=6
`therefore y= (6 -2x)/3`
`therefore` The point of the line can be taken as P =`( x,(6-2x)/3)`
Co -ordinate of origin are O = (0,0)
Co-ordinates of origin are (0,0)= O
`OP = sqrt((x-0)^2 + ((6-2x)/3 -0))^2`
∴ OP = `sqrt(x^2 + (2- (2x)/3)^2`
∴ `OP^2 = x^2 + (4x^2)/9+4-(8x)/3`
`OP^2 = (13x^2)/9 - (8x)/3 + 4`
Let 2 ( ), OP f x OP is minimum when OP2 is minimum.
`therefore f(x) = (13x^2)/9 - (8x)/3 + 4` ....(1)
Differentiating w.r.t. ‘x’ we get,
`f'(x)= 26/9>0`
`therefore` OP is minimum at `x = 12/13`
`therefore y = 2-(2x)/3`
`therefore y = y - 2/3xx12/13 = 2- 8/13= (26-8)/13 `
`therefore y=18/13`
∴ The closest point on the line 2 3 6 x y with origin is `(12/13 18/13)`
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