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In particular, for three points in the plane (n = 2), the above matrix is square and the points are collinear if and only if its determinant is zero; since that 3 × 3 determinant is plus or minus twice the area of a triangle with those three points as vertices, this is equivalent to the statement that the three points are collinear if and only ...
The equations originate from the central projection of a point of the object through the optical centre of the camera to the image on the sensor plane. [1] The three points P, Q and R are projected on the plane S through the projection centre C x- and z-axis of the projection of P through the projection centre C
By extension, k points in a plane are collinear if and only if any (k–1) pairs of points have the same pairwise slopes. In Euclidean geometry, the Euclidean distance d(a,b) between two points a and b may be used to express the collinearity between three points by: [3] [4]
They proved that the maximum number of points in the grid with no three points collinear is (). Similarly to Erdős's 2D construction, this can be accomplished by using points ( x , y , x 2 + y 2 {\displaystyle (x,y,x^{2}+y^{2}} mod p ) {\displaystyle p)} , where p {\displaystyle p} is a prime congruent to 3 mod 4 . [ 20 ]
If the three points are collinear, R can be informally considered to be +∞, and it makes rigorous sense to define c(x, y, z) = 0. If any of the points x, y and z are coincident, again define c(x, y, z) = 0. Using the well-known formula relating the side lengths of a triangle to its area, it follows that
In projective geometry, the harmonic conjugate point of a point on the real projective line with respect to two other points is defined by the following construction: Given three collinear points A, B, C, let L be a point not lying on their join and let any line through C meet LA, LB at M, N respectively.
Points that are incident with the same line are said to be collinear. The set of all points incident with the same line is called a range. If P 1 = (x 1, y 1, z 1), P 2 = (x 2, y 2, z 2), and P 3 = (x 3, y 3, z 3), then these points are collinear if and only if
Thus (E, H; J, G) = (E, K; D, L), so by Lemma X, the points H, M, and K are collinear. That is, the points of intersection of the pairs of opposite sides of the hexagon ADEGBZ are collinear. Lemmas XV and XVII are that, if the point M is determined as the intersection of HK and BG, then the points A, M, and D are collinear.