Point of intersection of pair of straight lines.












1














Let non homogenous curve equation be $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $ then first time partial differentiating with respect to x gives $ ax + hy + g =0$ and second time partial differentiating with respect to y gives $by + hx + f =0$ solving both equations give point of intersection of pair of lines. $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$



My question is - why partial differentiating makes work easy and what those two equations represent?






partial-derivative curves







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  • See math.stackexchange.com/q/1941210/265466 for several explanations.
    – amd
    Feb 9 '17 at 18:58
















1














Let non homogenous curve equation be $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $ then first time partial differentiating with respect to x gives $ ax + hy + g =0$ and second time partial differentiating with respect to y gives $by + hx + f =0$ solving both equations give point of intersection of pair of lines. $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$



My question is - why partial differentiating makes work easy and what those two equations represent?






partial-derivative curves







share|cite|improve this question
























  • See math.stackexchange.com/q/1941210/265466 for several explanations.
    – amd
    Feb 9 '17 at 18:58














1












1








1


1





Let non homogenous curve equation be $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $ then first time partial differentiating with respect to x gives $ ax + hy + g =0$ and second time partial differentiating with respect to y gives $by + hx + f =0$ solving both equations give point of intersection of pair of lines. $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$



My question is - why partial differentiating makes work easy and what those two equations represent?






partial-derivative curves







share|cite|improve this question















Let non homogenous curve equation be $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $ then first time partial differentiating with respect to x gives $ ax + hy + g =0$ and second time partial differentiating with respect to y gives $by + hx + f =0$ solving both equations give point of intersection of pair of lines. $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$



My question is - why partial differentiating makes work easy and what those two equations represent?






partial-derivative curves




partial-derivative curves






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 9 '17 at 16:16









amWhy

192k28225439




192k28225439










asked Feb 9 '17 at 15:10







user402003



















  • See math.stackexchange.com/q/1941210/265466 for several explanations.
    – amd
    Feb 9 '17 at 18:58


















  • See math.stackexchange.com/q/1941210/265466 for several explanations.
    – amd
    Feb 9 '17 at 18:58
















See math.stackexchange.com/q/1941210/265466 for several explanations.
– amd
Feb 9 '17 at 18:58




See math.stackexchange.com/q/1941210/265466 for several explanations.
– amd
Feb 9 '17 at 18:58










2 Answers
2






active

oldest

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0














You have some errors in your partial derivatives:



We start with your original equation: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $$



$(1)$ Finding the partial derivative with respect to $x$ gives: $$ax +hy +g=0$$



$(2)$ Finding the partial derivative with respect to $y$ gives $$2hx+2by+2f = by +hx+ f = 0$$ So in the first case, w.r.t. x, your $b$ in $by$ needs to be an $h$. In the second partial, w.r.t. y, your $us$ should be $hx.$



That gives us the equations of two lines:



$$begin{align} ax + hy+g &= 0 \ \
hx+by +f &= 0end{align}$$



Now we have a system of two equations and two variables, with $a, b, c, f, g, h$ all real numbers, from which we seek the solution $(x, y)$ at which the two given lines intersect.



Indeed, sollving the system gives us the intersection of the partial derivatives at $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$ provided $ab-h^2 leq 0$ and $bneq 0$.






share|cite|improve this answer























  • The question still remains unanswered : My question is - why partial differentiating makes work easy and what those two equations represent?
    – Yash Kumar Verma
    Sep 19 '18 at 20:11



















0














Suppose the lines are $L_1: ax+by+c=0$ and $L_2: dx+ey+f=0$. Then equation of pair of straight lines is $L_1 L_2 =0$. Now on partial differentiation w.r.t. $x$ and $y$ we get the following



$$aL_2+dL_1=0,,, text{ and } bL_2+eL_1=0.$$



Now using concept of linear algebra we know that these two equations have same solution as the original equations $L_1=0$ and $L_2=0$. (we can verify it easily). Hence solving the equations obtained on partial differentiation will give the point of intersection of two lines.



Hope that answers your question well.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    You have some errors in your partial derivatives:



    We start with your original equation: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $$



    $(1)$ Finding the partial derivative with respect to $x$ gives: $$ax +hy +g=0$$



    $(2)$ Finding the partial derivative with respect to $y$ gives $$2hx+2by+2f = by +hx+ f = 0$$ So in the first case, w.r.t. x, your $b$ in $by$ needs to be an $h$. In the second partial, w.r.t. y, your $us$ should be $hx.$



    That gives us the equations of two lines:



    $$begin{align} ax + hy+g &= 0 \ \
    hx+by +f &= 0end{align}$$



    Now we have a system of two equations and two variables, with $a, b, c, f, g, h$ all real numbers, from which we seek the solution $(x, y)$ at which the two given lines intersect.



    Indeed, sollving the system gives us the intersection of the partial derivatives at $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$ provided $ab-h^2 leq 0$ and $bneq 0$.






    share|cite|improve this answer























    • The question still remains unanswered : My question is - why partial differentiating makes work easy and what those two equations represent?
      – Yash Kumar Verma
      Sep 19 '18 at 20:11
















    0














    You have some errors in your partial derivatives:



    We start with your original equation: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $$



    $(1)$ Finding the partial derivative with respect to $x$ gives: $$ax +hy +g=0$$



    $(2)$ Finding the partial derivative with respect to $y$ gives $$2hx+2by+2f = by +hx+ f = 0$$ So in the first case, w.r.t. x, your $b$ in $by$ needs to be an $h$. In the second partial, w.r.t. y, your $us$ should be $hx.$



    That gives us the equations of two lines:



    $$begin{align} ax + hy+g &= 0 \ \
    hx+by +f &= 0end{align}$$



    Now we have a system of two equations and two variables, with $a, b, c, f, g, h$ all real numbers, from which we seek the solution $(x, y)$ at which the two given lines intersect.



    Indeed, sollving the system gives us the intersection of the partial derivatives at $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$ provided $ab-h^2 leq 0$ and $bneq 0$.






    share|cite|improve this answer























    • The question still remains unanswered : My question is - why partial differentiating makes work easy and what those two equations represent?
      – Yash Kumar Verma
      Sep 19 '18 at 20:11














    0












    0








    0






    You have some errors in your partial derivatives:



    We start with your original equation: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $$



    $(1)$ Finding the partial derivative with respect to $x$ gives: $$ax +hy +g=0$$



    $(2)$ Finding the partial derivative with respect to $y$ gives $$2hx+2by+2f = by +hx+ f = 0$$ So in the first case, w.r.t. x, your $b$ in $by$ needs to be an $h$. In the second partial, w.r.t. y, your $us$ should be $hx.$



    That gives us the equations of two lines:



    $$begin{align} ax + hy+g &= 0 \ \
    hx+by +f &= 0end{align}$$



    Now we have a system of two equations and two variables, with $a, b, c, f, g, h$ all real numbers, from which we seek the solution $(x, y)$ at which the two given lines intersect.



    Indeed, sollving the system gives us the intersection of the partial derivatives at $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$ provided $ab-h^2 leq 0$ and $bneq 0$.






    share|cite|improve this answer














    You have some errors in your partial derivatives:



    We start with your original equation: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $$



    $(1)$ Finding the partial derivative with respect to $x$ gives: $$ax +hy +g=0$$



    $(2)$ Finding the partial derivative with respect to $y$ gives $$2hx+2by+2f = by +hx+ f = 0$$ So in the first case, w.r.t. x, your $b$ in $by$ needs to be an $h$. In the second partial, w.r.t. y, your $us$ should be $hx.$



    That gives us the equations of two lines:



    $$begin{align} ax + hy+g &= 0 \ \
    hx+by +f &= 0end{align}$$



    Now we have a system of two equations and two variables, with $a, b, c, f, g, h$ all real numbers, from which we seek the solution $(x, y)$ at which the two given lines intersect.



    Indeed, sollving the system gives us the intersection of the partial derivatives at $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$ provided $ab-h^2 leq 0$ and $bneq 0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Feb 9 '17 at 16:43

























    answered Feb 9 '17 at 15:49









    amWhyamWhy

    192k28225439




    192k28225439












    • The question still remains unanswered : My question is - why partial differentiating makes work easy and what those two equations represent?
      – Yash Kumar Verma
      Sep 19 '18 at 20:11


















    • The question still remains unanswered : My question is - why partial differentiating makes work easy and what those two equations represent?
      – Yash Kumar Verma
      Sep 19 '18 at 20:11
















    The question still remains unanswered : My question is - why partial differentiating makes work easy and what those two equations represent?
    – Yash Kumar Verma
    Sep 19 '18 at 20:11




    The question still remains unanswered : My question is - why partial differentiating makes work easy and what those two equations represent?
    – Yash Kumar Verma
    Sep 19 '18 at 20:11











    0














    Suppose the lines are $L_1: ax+by+c=0$ and $L_2: dx+ey+f=0$. Then equation of pair of straight lines is $L_1 L_2 =0$. Now on partial differentiation w.r.t. $x$ and $y$ we get the following



    $$aL_2+dL_1=0,,, text{ and } bL_2+eL_1=0.$$



    Now using concept of linear algebra we know that these two equations have same solution as the original equations $L_1=0$ and $L_2=0$. (we can verify it easily). Hence solving the equations obtained on partial differentiation will give the point of intersection of two lines.



    Hope that answers your question well.






    share|cite|improve this answer


























      0














      Suppose the lines are $L_1: ax+by+c=0$ and $L_2: dx+ey+f=0$. Then equation of pair of straight lines is $L_1 L_2 =0$. Now on partial differentiation w.r.t. $x$ and $y$ we get the following



      $$aL_2+dL_1=0,,, text{ and } bL_2+eL_1=0.$$



      Now using concept of linear algebra we know that these two equations have same solution as the original equations $L_1=0$ and $L_2=0$. (we can verify it easily). Hence solving the equations obtained on partial differentiation will give the point of intersection of two lines.



      Hope that answers your question well.






      share|cite|improve this answer
























        0












        0








        0






        Suppose the lines are $L_1: ax+by+c=0$ and $L_2: dx+ey+f=0$. Then equation of pair of straight lines is $L_1 L_2 =0$. Now on partial differentiation w.r.t. $x$ and $y$ we get the following



        $$aL_2+dL_1=0,,, text{ and } bL_2+eL_1=0.$$



        Now using concept of linear algebra we know that these two equations have same solution as the original equations $L_1=0$ and $L_2=0$. (we can verify it easily). Hence solving the equations obtained on partial differentiation will give the point of intersection of two lines.



        Hope that answers your question well.






        share|cite|improve this answer












        Suppose the lines are $L_1: ax+by+c=0$ and $L_2: dx+ey+f=0$. Then equation of pair of straight lines is $L_1 L_2 =0$. Now on partial differentiation w.r.t. $x$ and $y$ we get the following



        $$aL_2+dL_1=0,,, text{ and } bL_2+eL_1=0.$$



        Now using concept of linear algebra we know that these two equations have same solution as the original equations $L_1=0$ and $L_2=0$. (we can verify it easily). Hence solving the equations obtained on partial differentiation will give the point of intersection of two lines.



        Hope that answers your question well.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 14:53









        M RizviM Rizvi

        163




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