Question

# Find the point on the line y=5x+3 that is closet to the origin.

Right triangles and trigonometry

Find the point on the line $$y=5x+3$$ that is closet to the origin.

2021-05-20

Let (x,y) be the point on the line that is closet to the origine. Since, $$y=5x+3,$$ so the point (x,y) on line becomes, $$(x,5x+3)$$ so now we use the distance dormula and minimize it.
Distance between (0,0) and $$(x,5x+3)$$ is given by
$$D=\sqrt{(x-0)^2+(5x+3-0)^2}$$
$$=\sqrt{x^2+(25x^2+30x+9)}$$
$$\Rightarrow D=\sqrt{26x^2+30x+9}$$
Now to minimize the distance, we find critical point by setting $$D'=0$$. Here using chain rule, derivative is:
$$D'=\frac{1}{2}(26x^2+30x+9)^{-\frac{1}{2}}\cdot\frac{d}{dx}(26x^2+30x+9)$$
$$=\frac{1}{2\sqrt{26x^2+30x+9}}\cdot(26\cdot2x+30\cdot1+0)$$
$$\Rightarrow D'=\frac{52x+30}{2\sqrt{26x^2+30x+9}}$$
Now D'=0
$$\Rightarrow \frac{52x+30}{2\sqrt{26x^2+30x+9}}=0$$
$$\Rightarrow 52x+30=0$$
$$\Rightarrow x=-\frac{30}{52}=-\frac{15}{26}$$