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. $\mathbf{S}(4,-2) ; m=\frac{1}{2}, \mathbf{T}(5,-3)$

   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.
$\mathbf{S}(4,-2) ; m=\frac{1}{2}, \mathbf{T}(5,-3)$
Modern Analytic Geometry
Modern Analytic Geometry
William Wooton,… 1st Edition
Chapter 3, Problem 9 ↓

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We have \(\mathbf{S}(4, -2)\) and \(\mathbf{T}(5, -3)\).  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. $\mathbf{S}(4,-2) ; m=\frac{1}{2}, \mathbf{T}(5,-3)$
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Key Concepts

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Distance from a Point to a Line Formula
This concept involves finding the shortest distance from a given point to a line. The mathematical procedure uses the formula |Ax? + By? + C| divided by the square root of A² + B², where Ax + By + C = 0 is the equation of the line and (x?, y?) are the coordinates of the point. It represents the perpendicular distance between the point and the line, which is the smallest possible distance.
Line Representation using a Point and a Slope
A line in a two-dimensional plane can be represented through a known point and its slope. This form is often expressed as y - y? = m(x - x?), where (x?, y?) is a point on the line and m is its slope. This representation is pivotal in converting the given slope into an equation that can be analyzed to determine distances or intersections.
Vector Projection and Orthogonal Distances
This concept refers to projecting one vector onto another to measure components along a direction orthogonal to a given line. In the context of finding the distance from a point to a line, the vector projection method helps in calculating the perpendicular (shortest) distance. The process involves finding the component of the vector connecting a known point on the line to the external point in the direction of the line's normal vector.

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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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