Question

find the distance between the given point $\mathbf{S}$ and the line having the given direction vector $v$ or slope $m$ and passing through the given point T. $\mathbf{S}(-3,-4) ; m=-\frac{1}{7}, \mathbf{T}(2,3)$ 

   find the distance between the given point $\mathbf{S}$ and the line having the given direction vector $v$ or slope $m$ and passing through the given point T.
$\mathbf{S}(-3,-4) ; m=-\frac{1}{7}, \mathbf{T}(2,3)$
Modern Analytic Geometry
Modern Analytic Geometry
William Wooton,… 1st Edition
Chapter 3, Problem 11 ↓

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We have \(\mathbf{S}(-3, -4)\) and \(\mathbf{T}(2, 3)\).  Show more…

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find the distance between the given point $\mathbf{S}$ and the line having the given direction vector $v$ or slope $m$ and passing through the given point T. $\mathbf{S}(-3,-4) ; m=-\frac{1}{7}, \mathbf{T}(2,3)$
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Key Concepts

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Line Representation in the Plane
This concept involves defining the equation of a line using a point-slope form, where a given point on the line and the slope (or a direction vector) determine the line uniquely. It is central to coordinate geometry and provides a way to describe a line algebraically, allowing for further geometric calculations.
Distance from a Point to a Line
This key concept refers to finding the shortest, perpendicular distance from a given point to a specified line. It is often derived from the point-to-line distance formula, which uses the coefficients from the line's equation and the coordinates of the point. This concept is essential for solving problems that involve limits, optimization, and analysis of geometric relationships.
Vector Projection and Orthogonality
When determining the distance from a point to a line, the idea of projecting one vector onto another is employed to find the component of the difference vector that is perpendicular to the line. This projection leverages the concept of orthogonality (perpendicularity) in vector spaces, providing insight into how to decompose vectors into components along specified directions.
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find the distance from the point to the line. (-1, 4, 3); x = 10 + 4t, y = -3, z = 4t
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