Problem 73
Question
Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (-1,-5) && 6 x+3 y=3\end{array}$$
Step-by-Step Solution
Verified Answer
The distance between the point (-1,-5) and the line \(6x+3y=3\) is \(\frac{8\sqrt{5}}{3}\)
1Step 1: Identify the variables in the formula
For the given line equation \(6x+3y=3\), a = 6, b = 3, and c = -3. For the given point (-1,-5), \(x_{1}=-1\) and \(y_{1}=-5\).
2Step 2: Substitute the variables into the formula
The distance formula now becomes: \[ d = \frac{| 6(-1) + 3(-5) - 3 |}{\sqrt{6^{2}+3^{2}}} \]
3Step 3: Simplify the equation
The numerators and denominators are to be simplified separately. Thus, the equation simplifies to: \[ d = \frac{| -6 -15 -3 |}{\sqrt{36+9}} \] and \[ d = \frac{| -24 |}{\sqrt{45}} \]
4Step 4: Calculate the distance
After simplification, the distance d becomes: \[ d = \frac{24}{\sqrt{45}} \] or \[d= \frac{8\sqrt{5}}{3}\]
Key Concepts
Distance FormulaEquation of a LineCoordinate GeometryAlgebraic Manipulation
Distance Formula
Understanding the distance formula is crucial in coordinate geometry. It provides a method to calculate the shortest distance between two entities in a plane. The distance between a point and a line is given by the formula: \
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
where \(a\), \(b\), and \(c\) are coefficients in the equation of the line in standard form \(ax + by + c = 0\), and \((x_1, y_1)\) represent the coordinates of the given point. The absolute value is used to ensure the distance is a non-negative value, and the denominator normalizes this distance, accounting for the coefficients' influence on the line's slope. By substituting the appropriate values into the formula, one can derive the shortest distance from the point to the line, which is always the perpendicular or orthogonal distance.
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
where \(a\), \(b\), and \(c\) are coefficients in the equation of the line in standard form \(ax + by + c = 0\), and \((x_1, y_1)\) represent the coordinates of the given point. The absolute value is used to ensure the distance is a non-negative value, and the denominator normalizes this distance, accounting for the coefficients' influence on the line's slope. By substituting the appropriate values into the formula, one can derive the shortest distance from the point to the line, which is always the perpendicular or orthogonal distance.
Equation of a Line
The equation of a line in the standard form \(ax + by + c = 0\) outlines a fundamental concept in coordinate geometry. Each letter represents a numerical coefficient: \(a\) and \(b\) correspond to the slope of the line, and \(c\) represents the y-intercept, which is the point where the line crosses the y-axis. To make use of the distance formula, it's necessary to express the line's equation in this standard form.
Understanding this can help when dealing with a variety of problems requiring an algebraic approach to geometrical issues. For example, if you start with the slope-intercept form of the equation (\(y = mx + b\)), you can rearrange it to match the standard form by algebraic manipulation, which is just manipulating the equation's terms through arithmetic operations to simplify or reorganize them.
Understanding this can help when dealing with a variety of problems requiring an algebraic approach to geometrical issues. For example, if you start with the slope-intercept form of the equation (\(y = mx + b\)), you can rearrange it to match the standard form by algebraic manipulation, which is just manipulating the equation's terms through arithmetic operations to simplify or reorganize them.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, merges algebra and geometry to describe the position of points, lines, and figures in a two-dimensional plane. The x and y coordinates define any point in this plane. The beauty of coordinate geometry lies in its ability to provide a bridge between algebraic equations and geometric shapes.
The distance between a point and a line, as seen in our exercise, is a classic problem that showcases the powerful combination of algebraic formulas and geometric concepts. Through coordinate geometry, complex geometric problems can be solved using algebraic methods, offering a clear and structured approach to these issues.
The distance between a point and a line, as seen in our exercise, is a classic problem that showcases the powerful combination of algebraic formulas and geometric concepts. Through coordinate geometry, complex geometric problems can be solved using algebraic methods, offering a clear and structured approach to these issues.
Algebraic Manipulation
Algebraic manipulation involves the techniques used to rearrange and simplify equations and expressions. It's a vital tool in mathematics, especially when dealing with equations of lines and distances in coordinate geometry. In our solution steps for finding the distance between a point and a line, the algebraic manipulation is used to substitute the values into the distance formula and then simplify the results.
For instance, expanding, factoring, and reducing expressions are all part of algebraic manipulation, which allows the transformation of complex equations into a more manageable form. It's these manipulations that enable the extraction of meaningful information, like the precise numerical distance between geometric entities.
For instance, expanding, factoring, and reducing expressions are all part of algebraic manipulation, which allows the transformation of complex equations into a more manageable form. It's these manipulations that enable the extraction of meaningful information, like the precise numerical distance between geometric entities.
Other exercises in this chapter
Problem 73
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Consider a hyperbola centered at the origin with a horizontal transverse axis. Use the definition of a hyperbola to derive its standard form.
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