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Specifically, stereographic projection from the north pole (0,1) onto the x-axis gives a one-to-one correspondence between the rational number points (x, y) on the unit circle (with y ≠ 1) and the rational points of the x-axis. If ( m / n , 0) is a rational point on the x-axis, then its inverse stereographic projection is the point
The coordinate surfaces of the Cartesian coordinates (x, y, z). The z-axis is vertical and the x-axis is highlighted in green. Thus, the red plane shows the points with x = 1, the blue plane shows the points with z = 1, and the yellow plane shows the points with y = −1.
Using point plotting, one associates an ordered pair of real numbers (x, y) with a point in the plane in a one-to-one manner. As a result, one obtains the 2-dimensional Cartesian coordinate system . To be able to plot points, one needs to first decide on a point in plane which will be called the origin , and a couple of perpendicular lines ...
A point in the plane may be represented in homogeneous coordinates by a triple (x, y, z) where x/z and y/z are the Cartesian coordinates of the point. [10] This introduces an "extra" coordinate since only two are needed to specify a point on the plane, but this system is useful in that it represents any point on the projective plane without the ...
The Smith chart graphical equivalent of using the transmission-line equation is to normalise , to plot the resulting point on a Z Smith chart and to draw a circle through that point centred at the Smith chart centre. The path along the arc of the circle represents how the impedance changes whilst moving along the transmission line.
Let the points on the circle be a sequence of coordinates of the vector to the point (in the usual basis). Points are numbered according to the order in which drawn, with n = 1 {\displaystyle n=1} assigned to the first point ( r , 0 ) {\displaystyle (r,0)} .
Now consider a point Q(x 1,0) and line segments PQ ⊥ OQ. The result is a right triangle OPQ with ∠QOP = t. Because PQ has length y 1, OQ length x 1, and OP has length 1 as a radius on the unit circle, sin(t) = y 1 and cos(t) = x 1. Having established these equivalences, take another radius OR from the origin to a point R(−x 1,y 1) on the ...
The tangent line through a point P on the circle is perpendicular to the diameter passing through P. If P = (x 1, y 1) and the circle has centre (a, b) and radius r, then the tangent line is perpendicular to the line from (a, b) to (x 1, y 1), so it has the form (x 1 − a)x + (y 1 – b)y = c.