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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
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If a point $(x,y)$ moves on a curve and satisfies the equation $\small\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\ax+by&bx+ay&0\end{vmatrix}=0$ then

$\begin{array}{1 1}(a)\;a,b,c\;form\;an\;AP\\(b)\;a,b,c\;form\;an\;HP\\(c)\;The\;point(x,y)\;lies \;on\;a\;curve\\(d)\;None\;of\;the\;above\end{array}$

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1 Answer

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Given
$\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\ax+by&bx+ay&0\end{vmatrix}=0$
Apply $C_3\rightarrow C_3-xC_1-yC_2$
$\Delta=\begin{vmatrix}a&b&0\\b&c&0\\ax+by&bx+ay&-(ax^2+ay^2+2bxy)\end{vmatrix}=0$
$\Rightarrow (b^2-ac)(ax^2+2bxy+ay^2)=0$
either $b^2=ac$
(or) $ax^2+2bxy+ay^2=0$
$\Rightarrow $The point $x,y$ lies on a curve through the origin.
Hence (c) is the right option
answered Nov 26, 2013 by sreemathi.v
edited Mar 20, 2014 by sharmaaparna1
 

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