/*! 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 23 Find the distance from the given... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the distance from the given point \(P\) to the given line \(L\). \(P(1,-1) ;\) Line \(L\) with equation \(4 x+5 y=1\)

Short Answer

Expert verified
The distance from point \(P\) to line \(L\) is \(\frac{2}{\sqrt{41}}\).

Step by step solution

01

Identify the values of A, B, and C

The equation of the given line \(L\) is \(4x + 5y = 1\). Comparing it to the standard form \(Ax + By + C = 0\), we can identify the values as: A = 4 B = 5 C = -1
02

Substitute the given point coordinates

We are given the point \(P(1, -1)\). So, \(x_0 = 1\) and \(y_0 = -1\). Now, substitute these values into the distance formula: \[Distance = \frac{|4(1) + 5(-1) -1|}{\sqrt{4^2 + 5^2}}\]
03

Calculate the distance

Now, we will calculate the distance using the formula: \[Distance = \frac{|4 - 5 - 1|}{\sqrt{16 + 25}}\] \[Distance = \frac{|-2|}{\sqrt{41}}\] \[Distance = \frac{2}{\sqrt{41}}\] So, the distance from point \(P\) to line \(L\) is \(\frac{2}{\sqrt{41}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Distance Formula
The distance from a point to a line in coordinate geometry can be calculated using a specific distance formula. This formula stems from the principle that the shortest distance between a point and a line is along the perpendicular drawn from the point to the line. To compute this distance, you employ the formula: \[Distance = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\] This formula requires that the equation of the line is in the standard form: \(Ax + By + C = 0\). When using the formula:
  • \(x_0\) and \(y_0\) represent the coordinates of the given point.
  • \(A\), \(B\), and \(C\) are the coefficients from the line equation.
The absolute value in the numerator ensures the distance remains positive, as distance is a scalar quantity. The denominator, consisting of the square root of the sum of squares of \(A\) and \(B\), scales the numerator to reflect the line’s orientation. By plugging values into the formula, you derive the exact distance from the point to the nearest part of the line.
Standard Form of a Line
The standard form of a line is a crucial concept in coordinate geometry and is written as \(Ax + By + C = 0\). It’s a way to express a linear equation where:
  • \(A\) and \(B\) are coefficients of \(x\) and \(y\), respectively.
  • \(C\) is a constant that shifts the line across the coordinate plane.
  • It’s essential for both \(A\) and \(B\) to be non-zero to represent a valid line.
  • \(A\), \(B\), and \(C\) are often integers for simplification.
This form is particularly useful because it allows for easy calculation of perpendicular distances, as illustrated in the distance formula. To convert any linear equation to this standard form, you must rearrange terms to conform to this structure. Ensuring the coefficients are integers can make solving problems simpler by avoiding fractional complexities. It also provides a consistent way to interpret and graph lines, serving as a foundation for more complex geometric analysis.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, merges algebra and geometry. It utilizes a coordinate system to establish relationships through algebraic equations. This branch of mathematics allows geometric problems to be expressed in terms of numerical equations, thus facilitating deeper analysis and systematic solutions. Key points of coordinate geometry include:
  • Positioning elements like points, lines, and shapes through coordinates.
  • Simplifying complex geometrical constructs to algebraic expressions.
  • Solving geometric problems using standard mathematical operations.
By using coordinates, it's easy to interpret the spatial relationships between different geometric entities. For example, identifying distances between points, slopes of lines, or intersections becomes methodical and precise in coordinate geometry. This tool is indispensable in various fields like engineering, physics, and computer graphics, as it provides a mathematical framework for construction and visualization in both two and three dimensions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A\) be an \(n \times n\) invertible matrix. Show that the unique solution to the linear system \(A \mathbf{x}=\mathbf{b},\) namely, \(\mathbf{x}=A^{-1} \mathbf{b},\) is also the least squares solution for this system.

Let \(\mathbf{v}=(7,-2) .\) Determine all nonzero vectors \(\mathbf{w}\) in \(\mathbb{R}^{2}\) such that \(\\{\mathbf{v}, \mathbf{w}\\}\) is an orthogonal set.

Suppose that a radioactive sample of a radioisotope decays exponentially according to the formula \(A(t)=\) \(A_{0} e^{k t}\) where \(k\) is a negative constant, \(A_{0}\) is the initial amount of the sample, \(t\) is measured in hours, and \(A(t)\) is measured in grams. Use a least squares technique for this nonlinear model to estimate a formula for \(A(t)\) given the following measured data: $$\begin{array}{c|c|c|c|cc}\hline \text { time } t \text { (in hours) } & 0 & 1 & 3 & 4 & 6 \\\\\hline \text { amount of sample } & 100 & 24.5 & 1.5 & 0.4 & 0.1 \\ A(t) \text { (in grams) } & & & & \\\\\hline\end{array}$$

If \(f(x)=\sin x\) and \(g(x)=2 \cos x+4\) on \([0, \pi / 2]\) use the function inner product ( 5.1 .5 ) to determine the angle between \(f\) and \(g\)

Let \(V\) be a real inner product space. (a) Prove that for all \(\mathbf{v}, \mathbf{w} \in V\) \(\|\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{v}\|^{2}+2\langle\mathbf{v}, \mathbf{w}\rangle+\|\mathbf{w}\|^{2}\) [Hint: \(\left.\|\mathbf{v}+\mathbf{w}\|^{2}=\langle\mathbf{v}+\mathbf{w}, \mathbf{v}+\mathbf{w}\rangle .\right]\) (b) Two vectors \(\mathbf{v}\) and \(\mathbf{w}\) in an inner product space \(V\) are called orthogonal if \(\langle\mathbf{v}, \mathbf{w}\rangle=0 .\) Use (a) to prove the general Pythagorean Theorem: If \(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal in an inner product space \(V,\) then $$\|\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.