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Find the equation of the pair of straight lines passing through the point $(2,1)$ and perpendicular to the pair of lines $4xy + 2x + 6y + 3$

The solution to the above problem is as such :

The required equation is $4(x-2)(y-1)$ $ xy-x-2y+2$

However I couldn’t understand how we can arrive at this solution ? Could you please explain how it’s done ? Thanks !

Aditi
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    If you mention an EQUAtion, it must possess an EQUAl sign. – Jean Marie Jan 05 '18 at 08:31
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    Looks like you might have made a transcription error. Your second expression, which I assume should be set equal to zero to make an equation, isn’t a pair of intersecting lines. Try plotting it. – amd Jan 05 '18 at 09:20

2 Answers2

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Note that the equation of the lines represented by the equation $$4xy + 2x + 6y +3=0$$ are given by $2x +3=0$ and $2y +1=0$.

Therefore you can find the equation of the lines perpendicular to these two lines and passing through $(2,1) $.

  • Yes I did reach that step but I’m very confused since the slope of one of the lines is zero and others’ is undefined so then how do I form the equation using $y-y_1 = m(x-x_1)$ since the line perpendicular to the line with zero slope will have infinite slope ! – Aditi Jan 05 '18 at 07:34
  • Then the required equations will be: $y-1 =0$ and $x-2=0$. The first one is perpendicular to $2x +3=0$ and the second to $2y+1=0$. –  Jan 05 '18 at 07:36
  • Multiplying both gives us: $$(x-2)(y-1)=0$$ $$\implies xy - x - 2y +2=0$$ our answer.. –  Jan 05 '18 at 07:37
  • Also there is no $4$ term in the answer.. –  Jan 05 '18 at 07:37
  • Adit - Line perpendicular to $y=k$ is $x=l$ right, and pair of straight lines is product of two linear equations – jonsno Jan 05 '18 at 07:37
  • @samjoe ohh thanks for the hint ! – Aditi Jan 05 '18 at 07:44
  • @Rohan thanks for explaining so clearly :) – Aditi Jan 05 '18 at 07:44
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If you find where the lines intersect, then you can transform the equation by translating, rotating, and then translating again. A pair of intersecting lines is a degenerate hyperbola, and there are several ways to find the center of a hyperbola from its general conic equation, such as the one discussed in this question. Applying that formula produces the point $\left(-\frac32,-\frac12\right)$.

Now, to transform the equation, first translate this point to the origin by making the substitutions $x\mapsto x-\frac32$, $y\mapsto y-\frac12$. Simplifying the resulting equation gives $4xy=0$. This looks plausible: if the pair of lines intersect at the origin, their equation shouldn’t have any $x$ or $y$ terms. We can drop the constant factor of $4$ at this point to make things simpler. (It should now be obvious that the original lines parallel the coordinate axes, but we don’t really need to know what the individual lines are.) Now, rotate by making the substitutions $x\mapsto-y$ and $y\mapsto x$. This changes the equation to $-xy=0$, but again, we can drop the factor of $-1$. Finally, translate the origin to the new intersection point via $x\mapsto x-2$, $y\mapsto y-1$ to get $(x-2)(y-1) = 0$, which expands into $xy-x-2y+2=0$.

amd
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