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:

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

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 $\left(\frac{1}{8465},\ \frac{92}{8465}\right)$