/*! 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 60 Find the distance between the po... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the distance between the point and the line. Point \((2,1)\) Line \(y=x+2\)

Short Answer

Expert verified
The distance between the point (2,1) and the line \(y = x + 2\) is \(\frac{1}{\sqrt{2}}\) or approximately 0.7071

Step by step solution

01

Reformat the line equation from slope-intercept to standard form

Line equation in slope-intercept form is \(y = x + 2\). Let's subtract \(x\) from both sides of the equation to get it into standard form \(ax + by - c = 0\). So the line equation becomes \(x - y + 2 = 0\). Therefore, \(a = 1\), \(b = -1\) and \(c = 2\).
02

Substitute values into the distance formula

Paint coordinates: \(x = 2\) and \(y = 1\). From Step 1, \(a = 1\), \(b = -1\) and \(c = 2\). Now, by using formula \(d = \frac{|ax + by - c|}{\sqrt{a^{2} + b^{2}}}\), we have \(d = \frac{| 1*2 + -1*1 - 2|}{\sqrt{1^{2} + -1^{2}}}\)
03

Evaluate the expression

Solve the equation: \(d = \frac{| 2 - 1 - 2|}{\sqrt{1 + 1}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}\). This is the exact form, and can be simplified to a decimal approximation

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point-Line Distance
In coordinate geometry, finding the distance between a point and a line is a common task that involves using the distance formula. The distance formula is helpful when you need to determine how far away a specific point is from a line. This is important in various applications, such as physics, engineering, and computer graphics. To find the distance between a point \(x_1, y_1\) and a line given by \ ax + by + c = 0 \, we use the formula:
  • \( d = \frac{| ax_1 + by_1 + c |}{\sqrt{a^2 + b^2}} \)
This formula calculates the perpendicular distance from a point to the line, which is the shortest distance possible. Perpendicular means a line that forms a right angle (90 degrees) with another line. When using this formula, remember:
  • The numerator \( | ax_1 + by_1 + c | \) represents the absolute value of substituting the point into the line equation.
  • The denominator \( \sqrt{a^2 + b^2} \) is simply the length of the vector \( (a, b) \), ensuring the right scale for the distance.
By dividing these, you get the shortest path from the point to the line.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, combines algebra and geometry using a coordinate system. It allows us to solve geometric problems by utilizing algebraic equations. This branch of mathematics is incredibly useful for visualizing and solving problems effectively by using graphs and coordinates.
In coordinate geometry:
  • Points are specified by ordered pairs \( (x, y) \), representing their position on the coordinate plane.
  • Lines are represented by equations that show the relationship between x and y coordinates, such as the slope-intercept form \( y = mx + c \) or the standard form \( ax + by + c = 0 \).
  • Shapes such as circles, parabolas, and ellipses can also be described with equations.
Coordinate geometry facilitates the analysis and understanding of geometric figures using equations, enabling us to compute distances, midpoints, slopes, and more.
Standard Form of a Line
The standard form of a line in coordinate geometry is one of the most common ways to express a linear equation. It is especially useful in calculating distances and intersections with other lines. The standard form of a line is expressed as:
  • \( ax + by + c = 0 \)
where \(a, b,\) and \(c\) are constants, and \(x\) and \(y\) are variables representing the coordinates on the plane.

Benefits of using standard form:
  • It clearly delineates the line equation and is simple to rearrange for various algebraic procedures.
  • In the context of the point-line distance, it directly plugs into the distance formula without any changes.
  • It is easier to work with when dealing with intercepts, as you can quickly rearrange the equation to find where the line crosses the axes.
To convert from slope-intercept form \( y = mx + c \) to standard form, you can simply rearrange the terms, ensuring \(a\) is positive for consistency and understanding. This forms the basis for solving numerous geometry problems efficiently.

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