Translation of axes

mj.alawami · #1 $Old$ June 22nd, 2009, 07:30 AM

For the equation,find it's new equation if the origin is translated to the new origin as indicated.
$x^2+y^2-8x-10y+12 ; O(1,1)$
Attempt:
$x=x'+1 , y=y'+1$
$(x+1)^2+(y+1)^2-8(x+1)-10(y+1)+12=0$
$=x^2+y^2-6x-8y-4=0$

Am I correct ?
Thank you

Soroban · #2 $Old$ June 22nd, 2009, 08:49 AM

Hello, mj.alawami!

Quote:

For the equation, find its new equation if the origin is translated to the new origin as indicated.

$x^2+y^2-8x-10y+12\:=\:0,\qquad O(1,1)$

Sorry, you are wrong . . .

For horizontal and vertical translations, we use the opposite signs.

. . $\begin{array}{ccc}(0,0) &\to & (1,1) \\ \hline \\[-4mm] x &\to& x - 1 \\ y &\to& y - 1 \end{array}$