Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Lines AB and CD are perpendicular. Points A, B, D can have any random coordinates and we know there value. Line CD can have any random length and we know its value. How can we calculate coordinates of point C ?

enter image description here

share|improve this question

migrated from stackoverflow.com Nov 17 '13 at 7:45

This question came from our site for professional and enthusiast programmers.

    
This question appears to be off-topic because it is about someone's geometry homework. –  talonmies Nov 17 '13 at 7:41
    
It is not home work. I am self educating myself and I don't have anyone to ask. –  vasili111 Nov 17 '13 at 7:43

2 Answers 2

up vote 0 down vote accepted

Suppose $A(a_1,a_2),B(b_1,b_2)$ and $D(d_1,d_2)$.

The line $(\epsilon_1)$ passing from these points has an equation of the form $Ax+By+C=0$.

Since $DC$ (lets call her $\epsilon_2$ ) is perpendicular to $\epsilon_1$ lets distinquish the following cases:

A. If $\epsilon_1$ has an equation of the form $x=x_0,x_0\in\mathbb{R}$. Then $\epsilon_2$ has an equation of the form $y=y_0,y_0\in\mathbb{R}$. You can easily find $y_0$ from $D$ and then find the coordinates of $C$ from the given length.

case 1

B. If $\epsilon_1$ has an equation of the form $y=y_0,y_0\in\mathbb{R}$. Then $\epsilon_2$ has an equation of the form $x=x_0,x_0\in\mathbb{R}$. You can easily find $x_0$ from $D$ and then find the coordinates of $C$ from the given length.

case 2

C. If $\epsilon_1$ isn't perpendicular to any of the axes $xx',yy'$ then $\epsilon_1$ has an equation of the form $y=ax+b$ (find $a,b$).Then, $\epsilon_2$ has an equation of the form $y-d_2=\lambda(x-d_1)$ where $\lambda\cdot a=-1$. The circle $C$ with center $D(d_1,d_2)$ and radius the given length $r$ has an equation of the form $(x-d_1)^2+(y-d_2)^2=r^2$. Solving the system of equations for $C$ and $\epsilon_2$ will give you the desired coordinates. (Notice: you'll find 2 points in each case!)

case 3

share|improve this answer

First of all, you provide too much information: We need only either A or B, let's say we use A and ignore B. Then you start by calculating the vector A->D: $$\mathbf v = A - D$$ Then calculate the unit vector in that direction: $$\mathbf e = \frac{\mathbf v}{|\mathbf v|}$$ The perpendicular vector can be obtained by exchanging the x and y components, and flipping the sign of one of them: $$\mathbf e'=\binom{-\mathbf e_y}{\mathbf e_x}$$ This vector $\mathbf e'$ points from $D$ to $C$. Finally, given the distance $l$ from $D$ to $C$, you can calculate $C$ as $$ C = D + l\mathbf e'$$ Actually, there is another solution: $$ C = D - l\mathbf e'$$

share|improve this answer
    
Strictly speaking, the vector A->D is $D-A$, not $A-D$. (But it gives the same two solutions, just in a different order.) –  TonyK Nov 17 '13 at 12:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.