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
















bumped to the homepage by Community♦ 7 hours ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.















  • 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

191k28224439




191k28224439










asked Feb 9 '17 at 15:10







user402003












bumped to the homepage by Community♦ 7 hours ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.







bumped to the homepage by Community♦ 7 hours ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.














  • 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










1 Answer
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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 at 20:11











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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 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 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









amWhy

191k28224439




191k28224439












  • 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 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 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 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 at 20:11


















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