Question

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

Right triangles and trigonometry
ANSWERED
asked 2021-05-19

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

Answers (1)

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
\(\Rightarrow \frac{52x+30}{2\sqrt{26x^2+30x+9}}=0\)
\(\Rightarrow 52x+30=0\)
\(\Rightarrow x=-\frac{30}{52}=-\frac{15}{26}\)

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