Prepend a vector with 90 degree angle to an existing one












0














First of all: my knowledge in mathematics is a bit rusty, so no matter how simple my question is, I afraid I in every case need a somewhat detailed answer:



I have a line from coordinates (x1,y1) to (x2,y2). Now I want to prepend a second line (x0,y0) to (x1,y1) before this one, but it should have a 90 degree angle to the first one.



Any idea how I can calculate my coordinates (x0,y0) so that the two lines form a right angle?



Finally all should be done in C programming language but I think this does not matter for this specific question.



Thanks for your patience!






geometry vectors angle







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    0














    First of all: my knowledge in mathematics is a bit rusty, so no matter how simple my question is, I afraid I in every case need a somewhat detailed answer:



    I have a line from coordinates (x1,y1) to (x2,y2). Now I want to prepend a second line (x0,y0) to (x1,y1) before this one, but it should have a 90 degree angle to the first one.



    Any idea how I can calculate my coordinates (x0,y0) so that the two lines form a right angle?



    Finally all should be done in C programming language but I think this does not matter for this specific question.



    Thanks for your patience!






    geometry vectors angle







    share|cite|improve this question

























      0












      0








      0







      First of all: my knowledge in mathematics is a bit rusty, so no matter how simple my question is, I afraid I in every case need a somewhat detailed answer:



      I have a line from coordinates (x1,y1) to (x2,y2). Now I want to prepend a second line (x0,y0) to (x1,y1) before this one, but it should have a 90 degree angle to the first one.



      Any idea how I can calculate my coordinates (x0,y0) so that the two lines form a right angle?



      Finally all should be done in C programming language but I think this does not matter for this specific question.



      Thanks for your patience!






      geometry vectors angle







      share|cite|improve this question













      First of all: my knowledge in mathematics is a bit rusty, so no matter how simple my question is, I afraid I in every case need a somewhat detailed answer:



      I have a line from coordinates (x1,y1) to (x2,y2). Now I want to prepend a second line (x0,y0) to (x1,y1) before this one, but it should have a 90 degree angle to the first one.



      Any idea how I can calculate my coordinates (x0,y0) so that the two lines form a right angle?



      Finally all should be done in C programming language but I think this does not matter for this specific question.



      Thanks for your patience!






      geometry vectors angle




      geometry vectors angle






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 23 '18 at 8:39









      Mike MaikaeferMike Maikaefer

      1




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          2 Answers
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          0














          When two lines are perpendicular, their slopes (denoted by $m$) are negative reciprocals. In general, you can calculate $m$ by using the following formula.



          $$m = frac{Delta y}{Delta x} = frac{y_2-y_1}{x_2-x_1}$$



          Call the slope of the first line $m_1$ and the slope of the second line $m_2$.



          $$m_1 = frac{y_2-y_1}{x_2-x_1}$$



          $$m_2 = frac{y_1-y_0}{x_1-x_0}$$



          Use



          $$m_1 = -frac{1}{m_2}$$



          which gives



          $$frac{y_2-y_1}{x_2-x_1} = -frac{1}{frac{y_1-y_0}{x_1-x_0}} implies frac{y_2-y_1}{x_2-x_1} = frac{x_1-x_0}{y_1-y_0}$$



          and you have $(x_1, y_1)$ and $(x_2, y_2)$, so you can reach an equation for $y_0$ and $x_0$ followed by calculating for $(x_0, y_0)$, for which there are infinite possibilities.






          share|cite|improve this answer























          • Ah ok...the length for the new line from (x0,y0) to (x1,y1) is predefined, so this would limit the infinite number of possibilites a bit?
            – Mike Maikaefer
            Nov 23 '18 at 8:58










          • If that's the case, the yes. If there is a fixed length, there would be only one solution which satisfies both the equation and the length given. (Two if you choose one coordinate on either side.) (If not, any $(x_0, y_0)$ satisfying the equation you get for $y_0$ and $x_0$ will work.)
            – KM101
            Nov 23 '18 at 9:03





















          0














          There's no need to deal with slopes at all. Let $(a, b)$ be a vector, then $J(a, b):=(-b,a)$ is the rotated vector $(a, b)$ by $90^circ$. Now let $p_i=(x_i,y_i)$. Then
          $$p_0=p_1+J(p_2-p_1),$$
          that is explicitly written
          $$begin{pmatrix}x_0\ y_0
          end{pmatrix}=begin{pmatrix}x_1\ y_1
          end{pmatrix}+Jbegin{pmatrix}x_2-x_1\y_2-y_1
          end{pmatrix}=begin{pmatrix}x_1\ y_1
          end{pmatrix}+begin{pmatrix}y_1-y_2\x_2-x_1
          end{pmatrix}=
          begin{pmatrix}x_1+y_1-y_2\ y_1+x_2-x_1
          end{pmatrix}.
          $$






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
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            active

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            0














            When two lines are perpendicular, their slopes (denoted by $m$) are negative reciprocals. In general, you can calculate $m$ by using the following formula.



            $$m = frac{Delta y}{Delta x} = frac{y_2-y_1}{x_2-x_1}$$



            Call the slope of the first line $m_1$ and the slope of the second line $m_2$.



            $$m_1 = frac{y_2-y_1}{x_2-x_1}$$



            $$m_2 = frac{y_1-y_0}{x_1-x_0}$$



            Use



            $$m_1 = -frac{1}{m_2}$$



            which gives



            $$frac{y_2-y_1}{x_2-x_1} = -frac{1}{frac{y_1-y_0}{x_1-x_0}} implies frac{y_2-y_1}{x_2-x_1} = frac{x_1-x_0}{y_1-y_0}$$



            and you have $(x_1, y_1)$ and $(x_2, y_2)$, so you can reach an equation for $y_0$ and $x_0$ followed by calculating for $(x_0, y_0)$, for which there are infinite possibilities.






            share|cite|improve this answer























            • Ah ok...the length for the new line from (x0,y0) to (x1,y1) is predefined, so this would limit the infinite number of possibilites a bit?
              – Mike Maikaefer
              Nov 23 '18 at 8:58










            • If that's the case, the yes. If there is a fixed length, there would be only one solution which satisfies both the equation and the length given. (Two if you choose one coordinate on either side.) (If not, any $(x_0, y_0)$ satisfying the equation you get for $y_0$ and $x_0$ will work.)
              – KM101
              Nov 23 '18 at 9:03


















            0














            When two lines are perpendicular, their slopes (denoted by $m$) are negative reciprocals. In general, you can calculate $m$ by using the following formula.



            $$m = frac{Delta y}{Delta x} = frac{y_2-y_1}{x_2-x_1}$$



            Call the slope of the first line $m_1$ and the slope of the second line $m_2$.



            $$m_1 = frac{y_2-y_1}{x_2-x_1}$$



            $$m_2 = frac{y_1-y_0}{x_1-x_0}$$



            Use



            $$m_1 = -frac{1}{m_2}$$



            which gives



            $$frac{y_2-y_1}{x_2-x_1} = -frac{1}{frac{y_1-y_0}{x_1-x_0}} implies frac{y_2-y_1}{x_2-x_1} = frac{x_1-x_0}{y_1-y_0}$$



            and you have $(x_1, y_1)$ and $(x_2, y_2)$, so you can reach an equation for $y_0$ and $x_0$ followed by calculating for $(x_0, y_0)$, for which there are infinite possibilities.






            share|cite|improve this answer























            • Ah ok...the length for the new line from (x0,y0) to (x1,y1) is predefined, so this would limit the infinite number of possibilites a bit?
              – Mike Maikaefer
              Nov 23 '18 at 8:58










            • If that's the case, the yes. If there is a fixed length, there would be only one solution which satisfies both the equation and the length given. (Two if you choose one coordinate on either side.) (If not, any $(x_0, y_0)$ satisfying the equation you get for $y_0$ and $x_0$ will work.)
              – KM101
              Nov 23 '18 at 9:03
















            0












            0








            0






            When two lines are perpendicular, their slopes (denoted by $m$) are negative reciprocals. In general, you can calculate $m$ by using the following formula.



            $$m = frac{Delta y}{Delta x} = frac{y_2-y_1}{x_2-x_1}$$



            Call the slope of the first line $m_1$ and the slope of the second line $m_2$.



            $$m_1 = frac{y_2-y_1}{x_2-x_1}$$



            $$m_2 = frac{y_1-y_0}{x_1-x_0}$$



            Use



            $$m_1 = -frac{1}{m_2}$$



            which gives



            $$frac{y_2-y_1}{x_2-x_1} = -frac{1}{frac{y_1-y_0}{x_1-x_0}} implies frac{y_2-y_1}{x_2-x_1} = frac{x_1-x_0}{y_1-y_0}$$



            and you have $(x_1, y_1)$ and $(x_2, y_2)$, so you can reach an equation for $y_0$ and $x_0$ followed by calculating for $(x_0, y_0)$, for which there are infinite possibilities.






            share|cite|improve this answer














            When two lines are perpendicular, their slopes (denoted by $m$) are negative reciprocals. In general, you can calculate $m$ by using the following formula.



            $$m = frac{Delta y}{Delta x} = frac{y_2-y_1}{x_2-x_1}$$



            Call the slope of the first line $m_1$ and the slope of the second line $m_2$.



            $$m_1 = frac{y_2-y_1}{x_2-x_1}$$



            $$m_2 = frac{y_1-y_0}{x_1-x_0}$$



            Use



            $$m_1 = -frac{1}{m_2}$$



            which gives



            $$frac{y_2-y_1}{x_2-x_1} = -frac{1}{frac{y_1-y_0}{x_1-x_0}} implies frac{y_2-y_1}{x_2-x_1} = frac{x_1-x_0}{y_1-y_0}$$



            and you have $(x_1, y_1)$ and $(x_2, y_2)$, so you can reach an equation for $y_0$ and $x_0$ followed by calculating for $(x_0, y_0)$, for which there are infinite possibilities.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 23 '18 at 8:58

























            answered Nov 23 '18 at 8:52









            KM101KM101

            5,7511423




            5,7511423












            • Ah ok...the length for the new line from (x0,y0) to (x1,y1) is predefined, so this would limit the infinite number of possibilites a bit?
              – Mike Maikaefer
              Nov 23 '18 at 8:58










            • If that's the case, the yes. If there is a fixed length, there would be only one solution which satisfies both the equation and the length given. (Two if you choose one coordinate on either side.) (If not, any $(x_0, y_0)$ satisfying the equation you get for $y_0$ and $x_0$ will work.)
              – KM101
              Nov 23 '18 at 9:03




















            • Ah ok...the length for the new line from (x0,y0) to (x1,y1) is predefined, so this would limit the infinite number of possibilites a bit?
              – Mike Maikaefer
              Nov 23 '18 at 8:58










            • If that's the case, the yes. If there is a fixed length, there would be only one solution which satisfies both the equation and the length given. (Two if you choose one coordinate on either side.) (If not, any $(x_0, y_0)$ satisfying the equation you get for $y_0$ and $x_0$ will work.)
              – KM101
              Nov 23 '18 at 9:03


















            Ah ok...the length for the new line from (x0,y0) to (x1,y1) is predefined, so this would limit the infinite number of possibilites a bit?
            – Mike Maikaefer
            Nov 23 '18 at 8:58




            Ah ok...the length for the new line from (x0,y0) to (x1,y1) is predefined, so this would limit the infinite number of possibilites a bit?
            – Mike Maikaefer
            Nov 23 '18 at 8:58












            If that's the case, the yes. If there is a fixed length, there would be only one solution which satisfies both the equation and the length given. (Two if you choose one coordinate on either side.) (If not, any $(x_0, y_0)$ satisfying the equation you get for $y_0$ and $x_0$ will work.)
            – KM101
            Nov 23 '18 at 9:03






            If that's the case, the yes. If there is a fixed length, there would be only one solution which satisfies both the equation and the length given. (Two if you choose one coordinate on either side.) (If not, any $(x_0, y_0)$ satisfying the equation you get for $y_0$ and $x_0$ will work.)
            – KM101
            Nov 23 '18 at 9:03













            0














            There's no need to deal with slopes at all. Let $(a, b)$ be a vector, then $J(a, b):=(-b,a)$ is the rotated vector $(a, b)$ by $90^circ$. Now let $p_i=(x_i,y_i)$. Then
            $$p_0=p_1+J(p_2-p_1),$$
            that is explicitly written
            $$begin{pmatrix}x_0\ y_0
            end{pmatrix}=begin{pmatrix}x_1\ y_1
            end{pmatrix}+Jbegin{pmatrix}x_2-x_1\y_2-y_1
            end{pmatrix}=begin{pmatrix}x_1\ y_1
            end{pmatrix}+begin{pmatrix}y_1-y_2\x_2-x_1
            end{pmatrix}=
            begin{pmatrix}x_1+y_1-y_2\ y_1+x_2-x_1
            end{pmatrix}.
            $$






            share|cite|improve this answer


























              0














              There's no need to deal with slopes at all. Let $(a, b)$ be a vector, then $J(a, b):=(-b,a)$ is the rotated vector $(a, b)$ by $90^circ$. Now let $p_i=(x_i,y_i)$. Then
              $$p_0=p_1+J(p_2-p_1),$$
              that is explicitly written
              $$begin{pmatrix}x_0\ y_0
              end{pmatrix}=begin{pmatrix}x_1\ y_1
              end{pmatrix}+Jbegin{pmatrix}x_2-x_1\y_2-y_1
              end{pmatrix}=begin{pmatrix}x_1\ y_1
              end{pmatrix}+begin{pmatrix}y_1-y_2\x_2-x_1
              end{pmatrix}=
              begin{pmatrix}x_1+y_1-y_2\ y_1+x_2-x_1
              end{pmatrix}.
              $$






              share|cite|improve this answer
























                0












                0








                0






                There's no need to deal with slopes at all. Let $(a, b)$ be a vector, then $J(a, b):=(-b,a)$ is the rotated vector $(a, b)$ by $90^circ$. Now let $p_i=(x_i,y_i)$. Then
                $$p_0=p_1+J(p_2-p_1),$$
                that is explicitly written
                $$begin{pmatrix}x_0\ y_0
                end{pmatrix}=begin{pmatrix}x_1\ y_1
                end{pmatrix}+Jbegin{pmatrix}x_2-x_1\y_2-y_1
                end{pmatrix}=begin{pmatrix}x_1\ y_1
                end{pmatrix}+begin{pmatrix}y_1-y_2\x_2-x_1
                end{pmatrix}=
                begin{pmatrix}x_1+y_1-y_2\ y_1+x_2-x_1
                end{pmatrix}.
                $$






                share|cite|improve this answer












                There's no need to deal with slopes at all. Let $(a, b)$ be a vector, then $J(a, b):=(-b,a)$ is the rotated vector $(a, b)$ by $90^circ$. Now let $p_i=(x_i,y_i)$. Then
                $$p_0=p_1+J(p_2-p_1),$$
                that is explicitly written
                $$begin{pmatrix}x_0\ y_0
                end{pmatrix}=begin{pmatrix}x_1\ y_1
                end{pmatrix}+Jbegin{pmatrix}x_2-x_1\y_2-y_1
                end{pmatrix}=begin{pmatrix}x_1\ y_1
                end{pmatrix}+begin{pmatrix}y_1-y_2\x_2-x_1
                end{pmatrix}=
                begin{pmatrix}x_1+y_1-y_2\ y_1+x_2-x_1
                end{pmatrix}.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 '18 at 11:45









                Michael HoppeMichael Hoppe

                10.8k31834




                10.8k31834






























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