Explained: "Find the point on the line y=5x+3 that..."

High School
Geometry
Trigonometry
Right triangles and trigonometry

Chesley

Answered question

2021-05-19

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

Answer & Explanation

mhalmantus

Skilled2021-05-20Added 105 answers

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

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