What is the minimum distance from the point (2 ,1) to the line y=2x? (do not approximate)...

Question:

What is the minimum distance from the point (2 ,1) to the line {eq}y=2x? {/eq} (do not approximate) (applications of derivatives)

Distance between Point and Line:

Suppose {eq}a x + b y + c = 0 {/eq} is an equation of the straight line where {eq}a {/eq} and {eq}b {/eq} are the coefficients of x and y respectively and {eq}(x_1, y_1) {/eq} a point on xy-plane, then the distance between the point and line is:

{eq}\displaystyle D = \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}} \ \ {/eq} where D is the distance between the point and line.

Answer and Explanation:


Given:

{eq}y = 2 x \text{ Point } (2, 1) {/eq}

We will compute the distance between the line and point.

First convert the given line in the standard form

{eq}2 x - y = 0 {/eq}

As we know the formula for the distance between the point and line is:

$$\begin{align*} \displaystyle D &= \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}} &\text{(Where } a = 2, b = - 1 ,c = 0, \text{ and } (x_1, y_1) = (2, 1) \text{)}\\ &= \frac{|2 \times 2 - 1 \times 1 + 0|}{\sqrt{2^2 + (-1)^2}} &\text{(Plugging in all given values)}\\ &= \frac{|4 - 1|}{\sqrt{4 + 1}} \\ &= \frac{|3|}{\sqrt{5}} \\ D \ &\boxed{= \frac{3}{\sqrt{5}} } \end{align*} $$


Learn more about this topic:

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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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