Search results
Results From The WOW.Com Content Network
The line with equation ax + by + c = 0 has slope -a/b, so any line perpendicular to it will have slope b/a (the negative reciprocal). Let ( m , n ) be the point of intersection of the line ax + by + c = 0 and the line perpendicular to it which passes through the point ( x 0 , y 0 ).
A linear equation in line coordinates has the form al + bm + c = 0, where a, b and c are constants. Suppose (l, m) is a line that satisfies this equation.If c is not 0 then lx + my + 1 = 0, where x = a/c and y = b/c, so every line satisfying the original equation passes through the point (x, y).
While the Lineweaver–Burk plot has historically been used for evaluation of the parameters, together with the alternative linear forms of the Michaelis–Menten equation such as the Hanes–Woolf plot or Eadie–Hofstee plot, all linearized forms of the Michaelis–Menten equation should be avoided to calculate the kinetic parameters ...
: distance from the origin of the line u {\displaystyle \mathbf {u} } : direction of line (a non-zero vector) Searching for points that are on the line and on the sphere means combining the equations and solving for d {\displaystyle d} , involving the dot product of vectors:
In analytic geometry, the intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line. It is the entire line if that line is embedded in the plane, and is the empty set if the line is parallel to the plane but outside it. Otherwise, the line cuts through the plane at a single point.
the distance between the two lines can be found by locating two points (one on each line) that lie on a common perpendicular to the parallel lines and calculating the distance between them. Since the lines have slope m, a common perpendicular would have slope −1/m and we can take the line with equation y = −x/m as a common perpendicular ...
The transfer time of a body moving between two points on a conic trajectory is a function only of the sum of the distances of the two points from the origin of the force, the linear distance between the points, and the semimajor axis of the conic. [2]
one solves the line equation for x or y and substitutes it into the equation of the circle and gets for the solution (using the formula of a quadratic equation) (,), (,) with x 1 / 2 = a c ± b r 2 ( a 2 + b 2 ) − c 2 a 2 + b 2 , {\displaystyle x_{1/2}={\frac {ac\pm b{\sqrt {r^{2}(a^{2}+b^{2})-c^{2}}}}{a^{2}+b^{2}}}\ ,}