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I want to calculate a point on a given line that is perpendicular from a given point.
I have a line segment AB and have a point C outside line segment. I want to calculate a point D on AB such that CD is perpendicular to AB.
I have to find point D.
It quite similar to this, but I want to consider to Z coordinate also as it does not show up correctly in 3D space.
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First, put the equation of the line given into slope-intercept form by solving for y. You get y = -2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6.
To find the perpendicular of a given line which also passes through a particular point (x, y), solve the equation y = (-1/m)x + b, substituting in the known values of m, x, and y to solve for b.
46,012. Lines which intersect each other at right angles are perpendicular to each other. In other words, if two lines are perpendicular to each other then, the angle made by them at the point of intersection will be 90o and vice versa.
Proof: Point D is on a line CD perpendicular to AB, and of course D belongs to AB. Write down the Dot product of the two vectors CD.AB = 0, and express the fact D belongs to AB as D=A+t(B-A).
We end up with 3 equations:
Dx=Ax+t(Bx-Ax) Dy=Ay+t(By-Ay) (Dx-Cx)(Bx-Ax)+(Dy-Cy)(By-Ay)=0 Subtitute the first two equations in the third one gives:
(Ax+t(Bx-Ax)-Cx)(Bx-Ax)+(Ay+t(By-Ay)-Cy)(By-Ay)=0 Distributing to solve for t gives:
(Ax-Cx)(Bx-Ax)+t(Bx-Ax)(Bx-Ax)+(Ay-Cy)(By-Ay)+t(By-Ay)(By-Ay)=0 which gives:
t= -[(Ax-Cx)(Bx-Ax)+(Ay-Cy)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2] getting rid of the negative signs:
t=[(Cx-Ax)(Bx-Ax)+(Cy-Ay)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2] Once you have t, you can figure out the coordinates for D from the first two equations.
Dx=Ax+t(Bx-Ax) Dy=Ay+t(By-Ay) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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