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Find the single equation of two straight lines that pass through the point $(2,3)$ and parallel to the line $x^2 - 6xy + 8y^2 = 0$.

My Attempt:

Let, $a_1x+b_1y=0$ and $a_2x+b_2y=0$ be the two lines represented by $x^2-6xy+8y^2=0$. then,

$$(a_1x+b_1y)(a_2x+b_2y)=0$$ $$(a_1a_2)x^2+(a_1b_2+b_1a_2)xy+(b_1b_2)y^2=0$$ Comparing with $x^2-6xy+8y^2=0$,

$a_1a_2=1, -(a_1b_2+b_1a_2)=6, b_1b_2=8$.

I got stuck at here. Please help me to continue this.

Rodrigo de Azevedo
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pi-π
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3 Answers3

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Note that $x^2-6xy+8y^2=(x-2y)(x-4y)$, so the lines represented by the equation $x^2-6xy+8y^2=0$ are $$ x-2y=0 \qquad \text{and} \qquad x - 4y = 0. $$ The line parallel to $x-2y=0$ and passing through $(2,3)$ is $$ (x-2) - 2(y-3) =0, $$ which in expanded form is $x - 2y + 4 = 0$. Similarly, the other line is $x-4y+10=0$. The single equation is then $$ (x - 2y + 4)(x - 4y + 10) = 0. $$

LaloVelasco
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  • How did you get $(x-2)-2(y-3)=0$? – pi-π Sep 17 '16 at 10:43
  • I just replaced $x$ by $x-2$ and $y$ by $y-3$ in the the ecuation $x-2y=0$. Think of it as a translation. You know that $x-2y=0$ is a line passing through the origin. Also, you want to obtain a straight line passing through $(2,3)$ but parallel to the given one. The geometric way of obtaining it is via a translation. You need to translate $(2,3)$ to $(0,0)$, so $(2,3)$ will be your new origin and the line will pass through it. How do we achieve a translation from $(2,3)$ to $(0,0)$? By mapping $(x,y)\mapsto(x,y) - (2,3)$, which is $(x-2,y-3)$. Then, we replace $x$ by $x-2$ and $y$ by $y-3$. – LaloVelasco Sep 17 '16 at 10:48
  • $x-4y+14=0$ does not pass through $(2,3)$. – mathlove Sep 17 '16 at 10:52
  • Sorry, it was $x-4y+10$. Thank you. – LaloVelasco Sep 18 '16 at 21:40
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$a_1a_2=1, -(a_1b_2+b_1a_2)=6, b_1b_2=8$.

I got stuck at here. Please help me to continue this.

We may set $a_1=a_2=1$ to have $$b_1+b_2=-6,\quad b_1b_2=8$$ giving $\{b_1,b_2\}=\{-2,-4\}$.

So, we get $$x^2-6xy+8y^2=(x-2y)(x-4y)$$ Hence, the equation we want is $$((x-2)-2(y-3))((x-2)-4(y-3))=0,$$ i.e. $$(x-2y+4)(x-4y+10)=0$$

Community
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mathlove
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The lines $$(x-2)^2 - 6(x-2)(y-3) + 8(y-3)^2 = 0$$ on transfer of origin to $(2,3)$, using the transformations $X = x-2, Y= y-2$ becomes $$X^2 - 6XY + 8Y^2 = 0$$ Hence the required equation is $$(x-2)^2 - 6(x-2)(y-3) + 8(y-3)^2 = 0$$

  • Is it necessary to take $-$ while transferring or can we take $+$, too? – pi-π Sep 17 '16 at 11:06
  • If we have an equation $f(x,y) = 0$, and transfer the origin to $(h,k)$, the new equation is $f(X+h, Y+k) = 0$. In the present case, the equation $(x-2)^2 - 6(x-2)(y-3) + 8(y-3)^2 = 0$ when the origin is transferred to $(2,3)$ becomes $(X+2-2)^2 - 6(X+2-2)(Y+3-3) + 8(Y+3-3)^2 = 0$ and hence $X^2 -6XY + 8Y^2 = 0$ with reference to the new axes. This is what we have used. –  Sep 17 '16 at 11:10
  • What does $x$ , $X$, $y$, $Y$ represent? – pi-π Sep 17 '16 at 11:16
  • $X, Y$ are the coordinates with respect to the new axes and $x,y$ are coordinates with respect to old axes. –  Sep 17 '16 at 14:32