For pair of st. lines , length of line joining feet of perpendiculars from $(f,g)$ to them is$sqrt {4.frac...












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Consider a pair of straight lines through the origin,
$$ax^2+2hxy+by^2=0$$
This can be written as,
$$y=m_{1,2}x$$
where $m_{1,2}=-frac {a}{h±sqrt {h^2-ab}}$. Now, suppose a point $(f,g)$ whence perpendiculars are drawn to both the lines . Then the eqations of the perpendiculars are,
$$y=-frac {1}{m_{1,2}}x + C_{1,2}$$
where $C_{1,2}=g-frac {h±sqrt {h^2-ab}}{a}f$. Finally , we can find the coordinates of the feet of perpendiculars using the formula of intersection coordinates:
$$(x,y)=(frac {c_1-c_2}{m_2-m_1},frac {c_1m_2-c_2m_1}{m_2-m_1})$$
Hence using the distance formula, the distance between the feet of perpendiculars is,
$$s=sqrt {(frac {C_2m_2}{1+m_2^2}-frac {C_1m_1}{1+m_1^2})^2+(frac {C_2m_2}{m_2+frac {1}{m_2}}-frac {C_1m_1}{m_1+frac {1}{m_1}})^2}$$
I am unable to proceed further to get the desired result i.e. $sqrt {4.frac {(h^2-ab)(f^2+g^2)}{(a-b)^2+4h^2}}$. Initially I felt the need of simplification, but that too didn't work , when even my simplified expression $s=sqrt {frac {(1+m_1^2)(f-gm_2)^2+(1+m_2^2)(f-gm_1)^2-2(f-gm_1)(f-gm_2)(1+m_1m_2)}{(1+m_1^2)(1+m_2^2)}}$ needed a CAS to be rendered in the desired form. Any help is welcome.






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    Consider a pair of straight lines through the origin,
    $$ax^2+2hxy+by^2=0$$
    This can be written as,
    $$y=m_{1,2}x$$
    where $m_{1,2}=-frac {a}{h±sqrt {h^2-ab}}$. Now, suppose a point $(f,g)$ whence perpendiculars are drawn to both the lines . Then the eqations of the perpendiculars are,
    $$y=-frac {1}{m_{1,2}}x + C_{1,2}$$
    where $C_{1,2}=g-frac {h±sqrt {h^2-ab}}{a}f$. Finally , we can find the coordinates of the feet of perpendiculars using the formula of intersection coordinates:
    $$(x,y)=(frac {c_1-c_2}{m_2-m_1},frac {c_1m_2-c_2m_1}{m_2-m_1})$$
    Hence using the distance formula, the distance between the feet of perpendiculars is,
    $$s=sqrt {(frac {C_2m_2}{1+m_2^2}-frac {C_1m_1}{1+m_1^2})^2+(frac {C_2m_2}{m_2+frac {1}{m_2}}-frac {C_1m_1}{m_1+frac {1}{m_1}})^2}$$
    I am unable to proceed further to get the desired result i.e. $sqrt {4.frac {(h^2-ab)(f^2+g^2)}{(a-b)^2+4h^2}}$. Initially I felt the need of simplification, but that too didn't work , when even my simplified expression $s=sqrt {frac {(1+m_1^2)(f-gm_2)^2+(1+m_2^2)(f-gm_1)^2-2(f-gm_1)(f-gm_2)(1+m_1m_2)}{(1+m_1^2)(1+m_2^2)}}$ needed a CAS to be rendered in the desired form. Any help is welcome.






    analytic-geometry







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      Consider a pair of straight lines through the origin,
      $$ax^2+2hxy+by^2=0$$
      This can be written as,
      $$y=m_{1,2}x$$
      where $m_{1,2}=-frac {a}{h±sqrt {h^2-ab}}$. Now, suppose a point $(f,g)$ whence perpendiculars are drawn to both the lines . Then the eqations of the perpendiculars are,
      $$y=-frac {1}{m_{1,2}}x + C_{1,2}$$
      where $C_{1,2}=g-frac {h±sqrt {h^2-ab}}{a}f$. Finally , we can find the coordinates of the feet of perpendiculars using the formula of intersection coordinates:
      $$(x,y)=(frac {c_1-c_2}{m_2-m_1},frac {c_1m_2-c_2m_1}{m_2-m_1})$$
      Hence using the distance formula, the distance between the feet of perpendiculars is,
      $$s=sqrt {(frac {C_2m_2}{1+m_2^2}-frac {C_1m_1}{1+m_1^2})^2+(frac {C_2m_2}{m_2+frac {1}{m_2}}-frac {C_1m_1}{m_1+frac {1}{m_1}})^2}$$
      I am unable to proceed further to get the desired result i.e. $sqrt {4.frac {(h^2-ab)(f^2+g^2)}{(a-b)^2+4h^2}}$. Initially I felt the need of simplification, but that too didn't work , when even my simplified expression $s=sqrt {frac {(1+m_1^2)(f-gm_2)^2+(1+m_2^2)(f-gm_1)^2-2(f-gm_1)(f-gm_2)(1+m_1m_2)}{(1+m_1^2)(1+m_2^2)}}$ needed a CAS to be rendered in the desired form. Any help is welcome.






      analytic-geometry







      share|cite|improve this question















      Consider a pair of straight lines through the origin,
      $$ax^2+2hxy+by^2=0$$
      This can be written as,
      $$y=m_{1,2}x$$
      where $m_{1,2}=-frac {a}{h±sqrt {h^2-ab}}$. Now, suppose a point $(f,g)$ whence perpendiculars are drawn to both the lines . Then the eqations of the perpendiculars are,
      $$y=-frac {1}{m_{1,2}}x + C_{1,2}$$
      where $C_{1,2}=g-frac {h±sqrt {h^2-ab}}{a}f$. Finally , we can find the coordinates of the feet of perpendiculars using the formula of intersection coordinates:
      $$(x,y)=(frac {c_1-c_2}{m_2-m_1},frac {c_1m_2-c_2m_1}{m_2-m_1})$$
      Hence using the distance formula, the distance between the feet of perpendiculars is,
      $$s=sqrt {(frac {C_2m_2}{1+m_2^2}-frac {C_1m_1}{1+m_1^2})^2+(frac {C_2m_2}{m_2+frac {1}{m_2}}-frac {C_1m_1}{m_1+frac {1}{m_1}})^2}$$
      I am unable to proceed further to get the desired result i.e. $sqrt {4.frac {(h^2-ab)(f^2+g^2)}{(a-b)^2+4h^2}}$. Initially I felt the need of simplification, but that too didn't work , when even my simplified expression $s=sqrt {frac {(1+m_1^2)(f-gm_2)^2+(1+m_2^2)(f-gm_1)^2-2(f-gm_1)(f-gm_2)(1+m_1m_2)}{(1+m_1^2)(1+m_2^2)}}$ needed a CAS to be rendered in the desired form. Any help is welcome.






      analytic-geometry




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      edited Nov 25 '18 at 10:14

























      asked Nov 25 '18 at 9:52









      Awe Kumar Jha

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          Write the lines like $(Ax+By)(A'x+B'y)=AA'x^2+2frac{AB'+A'B}{2}xy+BB'y^2=0$



          Then the system of one line and its perpendicular through $(f,g)$ is $langle Ax+By,Ay-Bx-Ag+Bfrangle$ and has solution $(x,y)=(frac{B}{A^2+B^2}(Bf-Ag),frac{A}{A^2+B^2}(Ag-Bf))$ and similarly for the primed, which gives the distance $$sqrt{frac{(AB'-A'B)^2(f^2+g^2)}{(A^2+B^2)(A'^2+B'^2)}}.$$ Now $h^2-ab=frac{(AB'-A'B)^2}{4}$ and $(a-b)^2+4h^2=(A^2+B^2)(A'^2+B'^2)$ and were done.






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            Write the lines like $(Ax+By)(A'x+B'y)=AA'x^2+2frac{AB'+A'B}{2}xy+BB'y^2=0$



            Then the system of one line and its perpendicular through $(f,g)$ is $langle Ax+By,Ay-Bx-Ag+Bfrangle$ and has solution $(x,y)=(frac{B}{A^2+B^2}(Bf-Ag),frac{A}{A^2+B^2}(Ag-Bf))$ and similarly for the primed, which gives the distance $$sqrt{frac{(AB'-A'B)^2(f^2+g^2)}{(A^2+B^2)(A'^2+B'^2)}}.$$ Now $h^2-ab=frac{(AB'-A'B)^2}{4}$ and $(a-b)^2+4h^2=(A^2+B^2)(A'^2+B'^2)$ and were done.






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              Write the lines like $(Ax+By)(A'x+B'y)=AA'x^2+2frac{AB'+A'B}{2}xy+BB'y^2=0$



              Then the system of one line and its perpendicular through $(f,g)$ is $langle Ax+By,Ay-Bx-Ag+Bfrangle$ and has solution $(x,y)=(frac{B}{A^2+B^2}(Bf-Ag),frac{A}{A^2+B^2}(Ag-Bf))$ and similarly for the primed, which gives the distance $$sqrt{frac{(AB'-A'B)^2(f^2+g^2)}{(A^2+B^2)(A'^2+B'^2)}}.$$ Now $h^2-ab=frac{(AB'-A'B)^2}{4}$ and $(a-b)^2+4h^2=(A^2+B^2)(A'^2+B'^2)$ and were done.






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                Write the lines like $(Ax+By)(A'x+B'y)=AA'x^2+2frac{AB'+A'B}{2}xy+BB'y^2=0$



                Then the system of one line and its perpendicular through $(f,g)$ is $langle Ax+By,Ay-Bx-Ag+Bfrangle$ and has solution $(x,y)=(frac{B}{A^2+B^2}(Bf-Ag),frac{A}{A^2+B^2}(Ag-Bf))$ and similarly for the primed, which gives the distance $$sqrt{frac{(AB'-A'B)^2(f^2+g^2)}{(A^2+B^2)(A'^2+B'^2)}}.$$ Now $h^2-ab=frac{(AB'-A'B)^2}{4}$ and $(a-b)^2+4h^2=(A^2+B^2)(A'^2+B'^2)$ and were done.






                share|cite|improve this answer












                Write the lines like $(Ax+By)(A'x+B'y)=AA'x^2+2frac{AB'+A'B}{2}xy+BB'y^2=0$



                Then the system of one line and its perpendicular through $(f,g)$ is $langle Ax+By,Ay-Bx-Ag+Bfrangle$ and has solution $(x,y)=(frac{B}{A^2+B^2}(Bf-Ag),frac{A}{A^2+B^2}(Ag-Bf))$ and similarly for the primed, which gives the distance $$sqrt{frac{(AB'-A'B)^2(f^2+g^2)}{(A^2+B^2)(A'^2+B'^2)}}.$$ Now $h^2-ab=frac{(AB'-A'B)^2}{4}$ and $(a-b)^2+4h^2=(A^2+B^2)(A'^2+B'^2)$ and were done.







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                answered Nov 25 '18 at 12:26









                Jan-Magnus Økland

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