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Section formula

The section formula in coordinate geometry is a fundamental mathematical tool used to calculate the coordinates of a point that divides the line segment joining two given points, either internally or externally, in a specified ratio m:n.^[1] This formula bridges algebraic expressions with geometric positions on a plane, enabling precise determination of point locations along a line.^[2] For internal division, where the point lies between the two endpoints A(x_1, y_1) and B(x_2, y_2), the coordinates of the dividing point P are given by
P\left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right). ^[3] In contrast, for external division, where the point lies outside the segment, the coordinates are
P\left( \frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n} \right), assuming m \neq n.^[1] A special case is the midpoint formula, derived when m:n = 1:1, yielding P\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).^[2] The section formula finds extensive applications in analytic geometry, such as computing centroids of triangles, verifying collinearity of points, and solving problems in vector geometry and computer graphics for interpolation between coordinates.^[3] It also aids in determining the ratio in which a given point divides a line segment, enhancing problem-solving in fields like physics for position vectors and engineering for spatial calculations.^[1]

Two-dimensional geometry

Internal division

In coordinate geometry, internal division occurs when a point lies between the two endpoints of a line segment and divides it in a specified ratio m:n, where m and n are positive real numbers indicating the proportional segments from the division point to each endpoint. This concept allows for the precise location of points along a line segment based on weighted proportions, forming a foundational tool in analytic geometry for solving problems involving positions and intersections.^[4] The coordinates of the point P that divides the line segment joining points A(x_1, y_1) and B(x_2, y_2) internally in the ratio m:n are given by the section formula:

P\left( \frac{n x_1 + m x_2}{m + n}, \frac{n y_1 + m y_2}{m + n} \right).

Here, the weights n and m reflect the relative distances, ensuring the point is positioned such that the segment from A to P is to the segment from P to B as m:n. This formula arises directly from the principles of coordinate systems in the plane.^[5]^[4] To derive the formula, consider the parametric representation along the line segment. The position of P can be expressed as a weighted average, where the coordinate is shifted from A by a fraction \frac{m}{m+n} of the total displacement to B. For the x-coordinate:

x = x_1 + \frac{m}{m + n} (x_2 - x_1) = x_1 + \frac{m x_2 - m x_1}{m + n} = \frac{(m + n) x_1 + m x_2 - m x_1}{m + n} = \frac{n x_1 + m x_2}{m + n}.

The y-coordinate follows analogously:

y = y_1 + \frac{m}{m + n} (y_2 - y_1) = \frac{n y_1 + m y_2}{m + n}.

This derivation relies on the proportionality of distances in a straight line, confirming the ratio through the uniform scaling of coordinates. A geometric proof using similar triangles reinforces this: construct horizontal lines from A and P to align with the y-level of B, and vertical lines to form triangles whose similarity (by AA criterion, sharing angles and proportional sides) yields the same ratio m:n in both x- and y-directions, leading to the formula.^[5]^[4] For example, if A(1, 2) and B(3, 4) are divided internally in the ratio $1:1, the coordinates of P are \left( \frac{1 \cdot 1 + 1 \cdot 3}{1 + 1}, \frac{1 \cdot 2 + 1 \cdot 4}{1 + 1} \right) = (2, 3), illustrating a balanced midpoint position.^[4] The origins of the section formula trace back to the development of analytic geometry by René Descartes in his 1637 work La Géométrie, where coordinate methods enabled algebraic representation of geometric divisions; it gained prominence in 19th-century textbooks as a standard tool for plane geometry problems.^[6]

External division

External division refers to the case where a point divides the line segment joining two points in a given ratio but lies outside the segment, extending the line beyond one of the endpoints.^[7] In coordinate geometry, this occurs when the division ratio m:n results in the point P being positioned such that the directed segments satisfy AP:PB = m:n, but with P not between A and B.^[3] For points A(x₁, y₁) and B(x₂, y₂), the coordinates of the point P dividing AB externally in the ratio m:n are given by:

P\left( \frac{n x_1 - m x_2}{n - m}, \frac{n y_1 - m y_2}{n - m} \right)

This formula can equivalently be written as \left( \frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n} \right), reflecting the sign convention for directed distances.^[8]^[3] The derivation of this formula relies on the geometry of similar triangles and the concept of directed segments. Consider points A(x₁, y₁) and B(x₂, y₂) on the plane, with P(x, y) dividing AB externally in ratio m:n. Drop perpendiculars from A, B, and P to the x-axis, meeting at points C, D, and M respectively, forming right triangles AMC and BND. Since P is external, the triangles ∆AMC and ∆BND are similar because their corresponding angles are equal (both right-angled, and sharing the angle with the line AB). The ratio of similarity is m:n, so the ratios of corresponding sides are equal: \frac{AM}{BN} = \frac{m}{n}. Substituting the lengths, AM = |x - x₁| and BN = |x₂ - x|, but accounting for direction in external division (where P is beyond B, say), yields \frac{x - x_1}{x_2 - x} = -\frac{m}{n} due to the opposite orientation. Solving \frac{x - x_1}{x_2 - x} = -\frac{m}{n}: cross-multiplying gives n(x - x_1) = -m (x_2 - x). Expanding, n x - n x_1 = -m x_2 + m x, then n x - m x = n x_1 - m x_2, so x (n - m) = n x_1 - m x_2, hence x = \frac{n x_1 - m x_2}{n - m}. A similar process for the y-coordinates, using vertical distances, yields y = \frac{n y_1 - m y_2}{n - m}. This proof emphasizes the opposite direction in external division compared to the internal case, where the ratio is positive without the sign flip.^[7]^[3] For example, consider points A(1, 2) and B(3, 4) divided externally in the ratio 2:1 (m=2, n=1). The coordinates of P are \left( \frac{1 \cdot 1 - 2 \cdot 3}{1 - 2}, \frac{1 \cdot 2 - 2 \cdot 4}{1 - 2} \right) = \left( \frac{1 - 6}{-1}, \frac{2 - 8}{-1} \right) = (5, 6). This point lies outside the segment AB, extended beyond B.^[8] The formula requires m ≠ n to avoid division by zero, which would occur if the ratio led to an indeterminate position; otherwise, the point would coincide with one endpoint or be undefined.^[7] This external counterpart to internal division finds applications in concepts like harmonic divisions, where points divide segments in specific negative ratios.^[3]

Midpoint formula

The midpoint of a line segment in two-dimensional coordinate geometry is defined as the point that divides the segment in the ratio 1:1, meaning it lies exactly halfway between the endpoints.^[9] This formula arises as a special case of the internal division section formula when the ratio m:n = 1:1. For endpoints A(x_1, y_1) and B(x_2, y_2), the coordinates of the midpoint M are given by

M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).

This expression simplifies the general section formula by averaging the coordinates directly, reflecting equal partitioning.^[1] A proof of the midpoint formula can be derived using vector averaging in coordinate geometry. Consider the position vectors of points A and B as \vec{A} = (x_1, y_1) and \vec{B} = (x_2, y_2). The midpoint M is the point whose position vector \vec{M} satisfies \vec{M} = \frac{\vec{A} + \vec{B}}{2}, which expands to the coordinate form above; this follows from the symmetry of the line segment and the property that the vector from A to M equals the vector from M to B in magnitude and direction opposition.^[10] Key properties of the midpoint include its role in the midpoint theorem for triangles, which states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long; this theorem implies the parallelogram law by demonstrating vector addition properties in geometric figures.^[11] For example, the midpoint of segment joining A(0,0) and B(4,6) is M(2,3), which geometrically represents the center of the segment and balances the coordinates equidistantly from both ends.^[12] The midpoint formula also aids in verifying segment properties via the distance formula, such as confirming that the distances from each endpoint to the midpoint are equal, establishing the bisection.^[13]

Special cases and applications

Centroid of a triangle

The centroid of a triangle is defined as the point of concurrency of its three medians, where each median connects a vertex to the midpoint of the opposite side and is divided by the centroid in the ratio 2:1, with the longer portion adjacent to the vertex.^[14] This point represents the geometric center of the triangle in terms of mass distribution when assuming equal masses at the vertices.^[15] In two-dimensional coordinate geometry, for a triangle with vertices A(x_1, y_1), B(x_2, y_2), and C(x_3, y_3), the coordinates of the centroid G are calculated as the average of the vertices' coordinates:

G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

This formula arises directly from the section formula applied to the medians. To derive it, consider the median from vertex A to the midpoint M of side BC, where M = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right). Since G divides segment AM in the ratio 2:1 (two parts toward M, one part toward A), the section formula gives:

G_x = \frac{2 \cdot \frac{x_2 + x_3}{2} + 1 \cdot x_1}{3} = \frac{x_1 + x_2 + x_3}{3}

The y-coordinate follows analogously. Repeating this for the other medians yields the same result, confirming concurrency at this point.^[14] Key properties of the centroid include its role as the balance point for a uniform triangular lamina, where it serves as the center of mass under constant density.^[15] It always lies inside the triangle, regardless of the triangle's shape.^[16] In barycentric coordinates, the centroid corresponds to the weighted average with equal weights of \frac{1}{3} for each vertex, reflecting its position as the arithmetic mean.^[17] Additionally, the centroid provides the reference for computing moments of inertia in triangular regions; for instance, the second moment of area about a centroidal axis parallel to the base establishes the minimum inertia value before applying the parallel axis theorem.^[18] As an example, consider a triangle with vertices at (0,0), (3,0), and (0,4). Applying the centroid formula yields G = \left( \frac{0 + 3 + 0}{3}, \frac{0 + 0 + 4}{3} \right) = \left(1, \frac{4}{3}\right), which lies inside the triangle and divides each median in the 2:1 ratio.^[14]

Section formula in conic sections

The section formula plays a crucial role in parametrizing points that divide chords of conic sections, enabling the determination of coordinates for intermediate points on curves such as ellipses, parabolas, and hyperbolas. In coordinate geometry, this involves applying the internal or external division formula to the parametric coordinates of chord endpoints, which lie on the conic. For a chord joining points (x_1, y_1) and (x_2, y_2) divided in the ratio m:n, the coordinates of the dividing point P are given by

x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}

for internal division, allowing precise location of points along focal chords or other segments within the conic. This parametrization is particularly useful in parabolas, where chords can be analyzed using the standard form y^2 = 4ax. Consider the latus rectum, a focal chord perpendicular to the axis with endpoints (a, 2a) and (a, -2a). The point dividing these endpoints in the ratio 1:1 (the midpoint) has coordinates (a, 0), which coincides with the focus of the parabola, illustrating how the section formula reveals key geometric properties. In ellipses, the section formula connects chord division to the eccentricity e. Specifically, the vertices of the ellipse can be derived from points dividing the segment between a focus and the corresponding point on the directrix in the ratio e:1. For an ellipse with focus S at (ae, 0) and directrix point Z at (a/e, 0), the points A and A' dividing SZ in this ratio yield the vertices at (a, 0) and (-a, 0), with the center as the midpoint, directly tying the division ratio to e = c/a where c is the focal distance.^[19] Applications extend to calculating tangents and normals at these division points. For a point P dividing a chord in a conic, the tangent at P can be found by substituting its coordinates into the conic's tangent equation derived from parametric forms; similarly, the normal follows from the derivative or perpendicular condition. In parametric equations for conics—such as (at^2, 2at) for parabolas or (a \cos \theta, b \sin \theta) for ellipses—the section formula integrates seamlessly to locate such points, facilitating derivations in modern analytic geometry. While Apollonius of Perga's ancient treatise Conics (c. 200 BCE) established foundational properties of these curves through synthetic methods, the section formula's coordinate-based approach, developed post-Cartesians, has enabled parametric analyses central to contemporary applications in physics and engineering.

Three-dimensional geometry

Internal and external division

In three-dimensional coordinate geometry, the section formula determines the coordinates of a point that divides the line segment joining two points A(x_1, y_1, z_1) and B(x_2, y_2, z_2) in a given ratio m:n, either internally (where the point lies between A and B) or externally (where the point lies outside the segment). This generalization incorporates the z-coordinate while preserving the proportional division principle from lower dimensions.^[20] For internal division in the ratio m:n (meaning the distances satisfy AP:PB = m:n), the coordinates of the point P are given by

P = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}, \frac{m z_2 + n z_1}{m + n} \right).

This formula arises from the parametric representation of the line through A and B. The parametric equations of the line are

\begin{align*} x &= x_1 + t (x_2 - x_1), \\ y &= y_1 + t (y_2 - y_1), \\ z &= z_1 + t (z_2 - z_1), \end{align*}

where t is a scalar parameter with t = 0 at A and t = 1 at B. For internal division, t = \frac{m}{m + n}, reflecting the fractional distance from A to P relative to AB. Substituting this value yields

x = x_1 + \frac{m}{m + n} (x_2 - x_1) = \frac{(m + n) x_1 + m x_2 - m x_1}{m + n} = \frac{n x_1 + m x_2}{m + n},

with analogous results for y and z.^[21]^[20] An alternative derivation relies on distance proportions in 3D space. Since A, P, and B are collinear, the Euclidean distances satisfy AP = \frac{m}{m + n} AB and PB = \frac{n}{m + n} AB. Projecting the segment onto the coordinate planes (XY, XZ, YZ) preserves collinearity and ratios due to orthogonal projections, leading to the same proportional coordinates in each dimension, analogous to similar triangles in planar geometry.^[22] For external division in the ratio m:n (where P lies on the extension of AB such that AP:PB = m:n with PB in the opposite direction), the coordinates are

P = \left( \frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n}, \frac{m z_2 - n z_1}{m - n} \right),

assuming m \neq n. In the parametric form, t = \frac{m}{m - n}, which can yield t > 1 or t < 0 depending on the ratio, placing P outside the segment. Substituting gives the negative sign in the numerator, as the effective contribution from B opposes that from A. The distance proportion proof follows similarly, with signed distances ensuring AP/PB = m/n while accounting for the external position.^[20]^[22] As an illustration, consider points A(1, 2, 3) and B(4, 5, 6) divided internally in the ratio 1:2. Here, m = 1, n = 2, so

P = \left( \frac{1 \cdot 4 + 2 \cdot 1}{3}, \frac{1 \cdot 5 + 2 \cdot 2}{3}, \frac{1 \cdot 6 + 2 \cdot 3}{3} \right) = (2, 3, 4).

This vector interpolation approach aligns with the abstract position vector formulation, where \vec{P} = \frac{n \vec{A} + m \vec{B}}{m + n} for internal division.^[20]

Midpoint and centroid in 3D

In three-dimensional space, the midpoint of a line segment joining two points A(x_1, y_1, z_1) and B(x_2, y_2, z_2) is the point that divides the segment in the ratio 1:1, given by the coordinates \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right).^[23] This formula extends the two-dimensional midpoint by incorporating the z-coordinate average, representing the geometric center of the segment where equal weights are balanced.^[24] For a tetrahedron with vertices A(x_1, y_1, z_1), B(x_2, y_2, z_2), C(x_3, y_3, z_3), and D(x_4, y_4, z_4), the centroid is the average of the vertex coordinates: \left( \frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + z_4}{4} \right).^[25] This point can be derived using the section formula applied to the medians of the tetrahedron, where each median joins a vertex to the centroid of the opposite face. The centroid divides each median in the ratio 3:1, with the longer segment from the vertex to the centroid.^[26] To see this, consider the median from vertex A to the face centroid G = \frac{B + C + D}{3}; applying the section formula for division in 3:1 yields \frac{1 \cdot A + 3 \cdot G}{4} = \frac{A + B + C + D}{4}, confirming the average as the intersection point of all medians.^[25] The centroid serves as the center of mass for a tetrahedron of uniform density, ensuring equilibrium when supported at that point, and it acts as the balance point in vector terms where the resultant force vector sums to zero.^[25] As an example, consider a cube with side length 1 and vertices at (0,0,0) and (1,1,1); the midpoint of this space diagonal is (0.5, 0.5, 0.5), which lies at the cube's center.^[27] For a tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), and (0,0,1), the centroid is at (0.25, 0.25, 0.25), equidistant in a balanced sense from the vertices.^[25] The centroid corresponds to the first moment of the mass distribution; higher moments, such as the second moment of inertia, measure distribution about this point and extend the concept to rotational dynamics.

Vector formulation

General section formula

The general section formula in vector notation offers a coordinate-free method to express the position vector of a point dividing the line segment joining two given points in a specified ratio. For internal division in the ratio m:n, where m and n are positive real numbers representing the proportional segments, the position vector \vec{P} of the dividing point P is

\vec{P} = \frac{n \vec{A} + m \vec{B}}{m + n},

with \vec{A} and \vec{B} denoting the position vectors of points A and B, respectively. This weighted average ensures P lies between A and B, with the weights n/(m+n) for \vec{A} and m/(m+n) for \vec{B} summing to unity.^[28] For external division in the same ratio m:n (assuming n > m > 0 to place P outside the segment on the side of A), the formula adjusts to account for the opposing direction:

\vec{P} = \frac{n \vec{A} - m \vec{B}}{n - m}.

Here, the negative sign reflects the extension beyond the segment, yielding weights that still sum to 1 but with one negative.^[28] These expressions derive from affine combinations of vectors, where the point is a linear blend with coefficients summing to 1, equivalent to barycentric weights along the line. To derive the internal case, parameterize the line from A to B such that the displacement to P is the fraction m/(m+n) of \vec{B} - \vec{A}: \vec{P} = \vec{A} + \frac{m}{m+n} (\vec{B} - \vec{A}). Expanding using vector addition and scalar multiplication properties yields \vec{P} = \frac{n \vec{A} + m \vec{B}}{m + n}. The external derivation follows analogously, with the parameter exceeding 1 or becoming negative, leading to the subtraction term.^[29]^[28] The vector formulation's primary advantages include invariance under rotations and translations, as position vectors shift rigidly without altering relative ratios, and straightforward extension to arbitrary dimensions beyond 2D or 3D coordinate systems.^[30] As an illustration, consider \vec{A} = \langle 1, 2 \rangle and \vec{B} = \langle 3, 4 \rangle dividing internally in the ratio 1:1. Then \vec{P} = \frac{1 \cdot \langle 1, 2 \rangle + 1 \cdot \langle 3, 4 \rangle}{2} = \langle 2, 3 \rangle. This approach underpins applications like the centroid as an equal-weight vector average.

Relation to coordinate systems

The vector formulation of the section formula aligns directly with its coordinate-based counterpart in Euclidean spaces, as position vectors are simply the coordinate representations of points relative to an origin. Specifically, when the components of the vector expression \vec{P} = \frac{n \vec{A} + m \vec{B}}{m + n} for internal division are extracted, they yield the standard coordinate formula P(x, y) = \left( \frac{n x_A + m x_B}{m + n}, \frac{n y_A + m y_B}{m + n} \right) in two dimensions, demonstrating their equivalence in Cartesian systems. This correspondence extends to three dimensions, where position vectors \langle x, y, z \rangle map precisely to rectangular coordinates, allowing seamless translation between vector operations and coordinate calculations.^[31]^[32] In higher dimensions, the section formula generalizes through barycentric coordinates, expressed as \vec{P} = \frac{\sum_{i=1}^k w_i \vec{A_i}}{\sum_{i=1}^k w_i}, where w_i are non-negative weights summing to 1 for points within a simplex in n-dimensional space; this affine combination enables the location of points dividing segments or simplices in arbitrary ratios across vector spaces of any dimension. Unlike the rigid two- or three-dimensional coordinate formulas, this vector approach scales naturally to n > 3, supporting computations in abstract vector spaces without rederiving component-wise equations. This formulation finds practical use in computer graphics through linear interpolation (lerp), where a point on a line segment is computed as \vec{P} = (1 - t) \vec{A} + t \vec{B} for $0 \leq t \leq 1, equivalent to internal division in ratio t : (1 - t), facilitating smooth transitions in rendering and animation. In physics, it underpins the center of mass calculation, \vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}, a weighted average that determines system dynamics independently of the origin. More recently, in machine learning, weighted averages of embeddings—such as \vec{e} = \sum w_i \vec{v}_i / \sum w_i, where \vec{v}_i are word vectors—generate document representations, enhancing semantic similarity tasks in models like Word2Vec.^[33]^[34]^[35] Vector formulations offer advantages over pure coordinate systems in non-Euclidean spaces, as they operate on tangent vectors at points, accommodating curved geometries like manifolds without assuming flat metric properties inherent to Cartesian grids. For instance, in hyperbolic or spherical spaces, vector-based weighted combinations preserve local linearity while coordinates would require metric-specific adjustments. To illustrate the equivalence in two dimensions, consider points A(1, 2) and B(5, 8) divided internally in the ratio 2:3; the coordinate formula gives P\left( \frac{2 \cdot 1 + 3 \cdot 5}{5}, \frac{2 \cdot 2 + 3 \cdot 8}{5} \right) = P(3.4, 5.6), matching the vector output \vec{P} = \frac{2 \vec{A} + 3 \vec{B}}{5} = \langle 3.4, 5.6 \rangle.^[32]

References

[1]
Section Formula - Definition, Formulas, Proof, Examples - Cuemath
The section formula is used to find the coordinates of the point that divides a line segment (externally or internally) into some ratio.
[2]
Proof for Sectional Formula - BYJU'S
Nov 10, 2020 · Section formula is used to determine the coordinate of a point that divides a line segment joining two points into two parts such that the ratio ...
[3]
Section Formula - GeeksforGeeks
Jul 23, 2025 · Section formula is used to find the coordinates of a point that divides the line in a given ratio or to find the ratio in which the line is divided by a point ...
[4]
[PDF] Coordinate geometry - Australian Mathematical Sciences Institute
Give an alternative proof of the previous theorem on internal division of an interval. Use the fact that AC = AP cosθ in the diagram from the proof, where θ ...
[5]
Proof of Internal Division Section formula - Math Doubts
Learn how to derive the internal division section formula geometrically to find the coordinates of a point that internally divides a line in a given ratio.Missing: derivation | Show results with:derivation
[6]
Descartes' Mathematics - Stanford Encyclopedia of Philosophy
Nov 28, 2011 · To speak of René Descartes' contributions to the history of mathematics is to speak of his La Géométrie (1637), a short tract included with ...
[7]
Section Formula: Internal & External Division, Proofs & Uses - Allen
In coordinate geometry, the section formula is an interesting tool that helps find a point that lies between any two locations, either internally or externally.
[8]
Section Formula | Brilliant Math & Science Wiki
The section formula tells us the coordinates of the point which divides a given line segment into two parts such that their lengths are in the ratio m : n.
[9]
Section Formula and Midpoint Formula - CK12-Foundation
A formula that will give you the coordinates of point based on the coordinates of , the coordinates of , and the ratio . This formula is sometimes referred to ...Missing: reliable | Show results with:reliable
[10]
Prove the midpoint formula using only the vector space axioms.
Aug 30, 2018 · The midpoint M of points A and B is given by the equation M−A=B−M. Prove that M=12(A+B) using only the vector space axioms.How to prove a formula for the midpoint of a line between two vectorsCoordinate geometry, proof of midpoint - Mathematics Stack ExchangeMore results from math.stackexchange.com
[11]
Midpoint Theorem - Statement, Proof, Converse, Examples - Cuemath
The midpoint theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the ...
[12]
Midpoint formula | Analytic geometry (article) - Khan Academy
The midpoint formula is basically an average. You add the two x-values and divide by 2. You add the two y-values and divide by 2. This gives you the coordinates ...Missing: sources | Show results with:sources
[13]
Coordinate Geometry Proofs - MathBitsNotebook(Geo)
Coordinate geometry proofs employ the use of formulas such as the Slope Formula, the Midpoint Formula and the Distance Formula, as well as postulates, theorems ...
[14]
[PDF] Coordinate Geometry
Sep 6, 2005 · Coordinate geometry is a subset of algebra, where plane geometry theorems can be formulated as polynomial equation systems. Transformation ...
[15]
Various points related to triangles
Mar 12, 2004 · The centroid of a triangle is the point at which the three medians meet. A median is the line between a vertex and the midpoint of the opposite ...
[16]
[PDF] Chapter 7.2-7.3 Triangle Centers and Regular Polygon Properties
Triangle centers include incenter, circumcenter, centroid, and orthocenter. Regular polygons have an incenter, radius, and apothem.Missing: derivation section
[17]
Barycentric coordinates
$$ p$ is at the centroid of the triangle, then $ \alpha_1 = \alpha_2 = \alpha_3 = 1/3$ . If $ p$ lies on an edge, then the opposing vertex weight is zero ...
[18]
2D Centroid and Moment of Inertia Table - Mechanics Map
Centroids and Area Moments of Inertia for 2D Shapes ; Triangle. Centroid of a Triangle Area=12bh A r e a = 1 2 b h, Ix=136bh3 I x = 1 36 b h 3. Ix′=112bh3 I x ′ ...Missing: derivation | Show results with:derivation
[19]
How to Find the Foci of an Ellipse?
Therefore, Foci = (ae, 0) & (-ae, 0) due to the symmetry. Let us consider that P(x,y) be the point on ellipse and PM be the perpendicular distance from P to l.
[20]
Section Formula in 3D - BYJU'S
Determine the coordinate points which divide the line segment that joins the point (1, -2, 3) and (3, 4, -5) in the ratio of 2:3 internally and externally.
[21]
Calculus III - Equations of Lines - Pauls Online Math Notes
Aug 15, 2023 · In this section we will derive the vector form and parametric form for the equation of lines in three dimensional space.
[22]
Section Formula in 3 Dimension Made Easy - Maths - Vedantu
Rating 4.2 (373,000) The derivation of the section formula in 3D is based on the concept of similar triangles and coordinate geometry. Imagine a line segment PQ in 3D space ...<|control11|><|separator|>
[23]
Geneseo Math 223 01 3D Coordinates - SUNY Geneseo
The idea of 3D coordinates. The distance formula in 3 dimensions, and other formulas such as the midpoint formula extended from 2 dimensions to more. Equations ...
[24]
11-01 3-D Coordinate System - Andrews University
Distance and Midpoint Formulas ... In two dimensions, the midpoint formula is the average or arithmetic mean of the two endpoints of the segment. This is still ...
[25]
[PDF] Computational Geometry Lab: TETRAHEDRONS
Dec 23, 2010 · The centroid is just the average of the vertices: Centroid = a + b + c + d 4 Here, of course, we mean that the x, y and z coordinates of the ...
[26]
[PDF] CLOSE YOUR EYES AND SEE… AN APPROACH TO SPATIAL ...
i) The four medians of F meet in its centroid. ii) The centroid of F divides each median in the ratio 1:3, the longer segment being on the side of the ...
[27]
[PDF] Geometry of Masses, Lesson 2 - UCLA Math Circle
Let M be the midpoint of the diagonal AD. To prove the statement, we need to show that the diagonal CB also passes through M and that |CM| = |MB|. 5. Page 6 ...
[28]
[PDF] Barycentric Coordinates: Formula Sheet
... internal division: R = (qu1 + pu2 q + p. , qv1 + pv2 q + p. , qw1 + pw2 q + p . ) . (12) and the formula for the external division: R = (qu1 - pu2 q - p. , qv1 ...
[29]
[PDF] Vectors - Web.math.wisc.edu
Given two distinct points A and B we consider the line segment AB. If X is any given point on AB then we will now find a formula for the position vector of X.
[30]
[PDF] Clifford Algebra to Geometric Calculus - MIT Mathematics
... coordinate-free differential calculus, and the whole subject can be regarded ... vector formulation while fitting it into the general theory of ...
[31]
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/12%3A_Vectors_in_Space/12.02%3A_Vectors_in_Three_Dimensions
[32]
Section Formula In Vector Algebra - BYJU'S
Sep 28, 2020 · The concept of section formula is implemented to find the coordinates of a point dividing a line segment internally or externally in a specific ...
[33]
Interpolation - Scratchapixel
a ( 1 − t ) + b t with 0 ≤ t ≤ 1 Figure 1: To compute the value at the grey point, we linearly interpolate the values a and b. The value t is in the range 0 to ...
[34]
9.9: Center of Mass (Part 1)
### Summary: Formula for Center of Mass in Vectors and Weighted Average
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Learning Word Embedding with Better Distance Weighting and Window Size Scheduling
### Summary: Weighted Averages in Word Embeddings for Machine Learning