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Given a line segment from A (10, 100) to B (300, 300);
and then point A' (10, 200) where A' has the same x, but a different y, as A, how do I find the point where a line from A' would intersect the line AB at D so that A, A', and D form a right triangle?
Get the equation for the line and substitute in the y you already have.
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This is a follow-up question to this question.
Taking the following image as an example:
What I know:
x and y coordinates of points D, E, and P.
Therefore, I also know slope and intercept of D-E line
What I want to know:
x and y coordinates of point Q. (This is the point which crosses the D-E line).
Notation P=[px,py], D=[dx,dy], E=[ex,ey], Q=[qx,qy]
First:
R=P-D=[px-dx, py-dy]=[rx,ry]
K=E-D=[ex-dx, ey-dy]=[kx, ky]
Then
z=dot(R,K)/dot(K,K)=(rx*kx+ry*ky) / (kx*kx+ky*ky)
Finally
Q=D+z*K=[dx+z*kx, dy+z*ky]
The R is vector which start on point D and ends on point K, the K is vector which start on point D and ends on point E. Using this we made scalar projection to calculate result Q. More info about concept here
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y=0 and x=0
Both increment by 1 every millisecond, but x=0 whenever it reaches 120.
What is the equation for finding x if y is... 5000?
x = y % 120
% is called modulus
You just use modulus.
(y / 120) + (y % 120) = x
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I want to rotate a plane represented by the equation z = 6 , by n degrees along y axis and find the new equation of the plane. how can this be done?
Thanks
Base point (0,0,6) after rotation will lie in XZ plane with coordinates
(x0, y0, z0) = (-6*sin(Fi), 0, 6*cos(Fi))
normal vector
n = (A,B,C) = (-sin(Fi), 0, cos(Fi))
so new plane equation is (for explanation see beginning of the article)
A*(x-x0)+B*(y-y0)+C*(z-z0)=0
or
-sin(Fi)*x + cos(Fi)*z - 6 = 0
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I know how to find orthocenter when coordinates of 3 vertices of a triangle is given. But what will be the coordinates of three vertex of the triangle in this question?
what does x=2 and y=3 means? does it mean A(2,0) and B(0,3) ??
x=2 is the line parallel to y axis and constant value x=2 ( perpendicular to x axis), similarly y=3 is line perpendicular to y axis , with constant y coordinate = 3 and parallel to x axis.
your answer :
see, basically what you are getting is an right angle triangle. Whose orthocentre is at 2,3 which is vertex of the triangle at the right angle. If you try to draw three lines given, you will get it. Hope it helps.
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how can we calculate the coordinates of a line (Formula used) when the length and angle is given.
I have to make a pop up box which will take as input the length and angle and will draw a line.
just need the formula
Assuming one end of the line is at (x0, y0), the other end will be at:
x1 = x0 + r cos(t * pi / 180)
y1 = y0 + r sin(t * pi / 180)
where r is the length of the line and t is the angle in degrees.