/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Find the point \(P\) on the line... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 19

Find the point \(P\) on the line \(y=3 x\) that is closest to the point \((50,0) .\) What is the least distance between \(P\) and (50,0)\(?\)

Short Answer

Expert verified
Answer: The least distance between the point (50,0) and the line y=3x is 15鈭�10.

Step by step solution

01

Find the slope of the line perpendicular to y=3x

The given line y=3x, so its slope, m1 = 3. The slope of the line perpendicular to this line, m2, should satisfy the condition m1*m2=-1. Thus, m2 = -\frac{1}{3}.
02

Write the equation of the perpendicular line

We will use the point-slope form, which is y-y1=m(x-x1), where m is the slope and (x1,y1) is the given point. Using the point (50,0) and the perpendicular slope -\frac{1}{3}, the equation becomes y-0=-\frac{1}{3}(x-50). Simplify this equation to obtain y=-\frac{1}{3}(x-50).
03

Find the intersection of the two lines to determine point P

We will find the intersection point of the lines y=3x and y=-\frac{1}{3}(x-50) by solving the system of equations. Equate the two y expressions and solve for x: 3x = -\frac{1}{3}(x-50) => 9x = -(x-50) => 10x = 50 => x = 5 Now substitute this x value back into y=3x to find the corresponding y value: y=3(5) => y=15 So, the point P is (5,15).
04

Calculate the distance between point P and (50,0) using the distance formula

Using the distance formula d=\sqrt{(x2-x1)^2+(y2-y1)^2}, where (x1,y1)=(50,0) and (x2,y2)=(5,15), we get: d = \sqrt{(5-50)^2+(15-0)^2}=\sqrt{(-45)^2+15^2}=\sqrt{2025+225}=\sqrt{2250} = 15\sqrt{10} Thus, the least distance between point P and (50,0) is 15\sqrt{10}.

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