# Find the point P on the line y=5x that is closest to the point (52,0). What is the least distance...

## Question:

Find the point {eq}P {/eq} on the line {eq}y=5x {/eq} that is closest to the point {eq}(52,0) {/eq}. What is the least distance between {eq}P {/eq} and {eq}(52,0) {/eq}?

## 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 defined as:

{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 = 5 x \text{ Point } (52, 0) {/eq}

We will compute the distance between the line and point.

First convert the given line in the standard form

{eq}5 x - y = 0 {/eq}

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

$$\begin{align*} \displaystyle D &= \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}} &\text{(Where } a = 5, b = - 1 ,c = 0, \text{ and } (x_1, y_1) = (52, 0) \text{)}\\ &= \frac{|5 \times 52 - 1 \times 0 + 0|}{\sqrt{5^2 + (-1)^2}} &\text{(Plugging in all given values)}\\ &= \frac{|260 -0 + 0|}{\sqrt{25 + 1}} \\ &= \frac{|260|}{\sqrt{26}} \\ &= \frac{260}{\sqrt{26}} \\ &= \frac{\sqrt{26} \sqrt{26} \times 10}{\sqrt{26}} \\ D \ &\boxed{= 10 \sqrt{26} } \end{align*} $$

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#### Learn more about this topic:

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

Chapter 30 / Lesson 5