For pair of st. lines , length of line joining feet of perpendiculars from $(f,g)$ to them is$sqrt {4.frac...
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
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
add a comment |
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
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
add a comment |
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
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
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
analytic-geometry
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
edited Nov 25 '18 at 10:14
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
asked Nov 25 '18 at 9:52
Awe Kumar Jha
38613
38613
add a comment |
add a comment |
1 Answer
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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.
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
add a comment |
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1 Answer
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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.
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
add a comment |
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.
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
add a comment |
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.
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
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.
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
answered Nov 25 '18 at 12:26
Jan-Magnus Økland
1,9111914
1,9111914
add a comment |
add a comment |
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