Question

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

Answer 1

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=√((x-1)^2+(92x-0)^2) \\ \\ =√(x^2-2x+1+8464x^2)=√(8465x^2-2x+1)`

For shortest distance,


`(dL)/(dx) =0`


`(dL)/(dx) =0 \\ \\ \Rightarrow (16930x-2)/(2√(8465x^2-2x+1)) =0 \\ \\ \Rightarrow16930x=2 \\ \\ \Rightarrow x= (1)/(8465)`

and
`y=92\left((1)/(8465)\right)=(92)/(8465)`

Therefore, the required point is
`\left((1)/(8465),\ (92)/(8465)\right)`