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aksik [14]
3 years ago
12

Find the point on the line y = 92x closest to the point (1, 0)

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
8 0
Let the point on the line y = 92x closest to the point (1, 0) be (x, 92x), then the distance between point (1, 0) and (x, 92x) is given by:

L=\sqrt{(x-1)^2+(92x-0)^2} \\  \\ =\sqrt{x^2-2x+1+8464x^2}=\sqrt{8465x^2-2x+1}

For shortest distance,

\frac{dL}{dx} =0

\frac{dL}{dx} =0 \\  \\ \Rightarrow  \frac{16930x-2}{2\sqrt{8465x^2-2x+1}} =0 \\  \\ \Rightarrow16930x=2 \\  \\ \Rightarrow x= \frac{1}{8465}

and y=92\left(\frac{1}{8465}\right)=\frac{92}{8465}

Therefore, the required point is \left(\frac{1}{8465},\ \frac{92}{8465}\right)
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