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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Show that the points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

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Toolbox:
  • The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
  • $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
  • $|\bigtriangleup|=\frac{1}{2}\begin{bmatrix}x_1\begin{vmatrix}y_2 & 1\\y_3 & 1\end{vmatrix}-y_1\begin{vmatrix}x_2 & 1\\x_3 & 1\end{vmatrix}+1\begin{vmatrix}x_2 & y_2\\x_3 & y_3\end{vmatrix}\end{bmatrix}$
Let $(a,(b+c)=(x_1,y_1)$
 
Let $(b,c+a)=(x_2,y_2)$
 
Let $(c,a+b)=(x_3,y_3)$
 
Now area of a triangle is given by
 
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}$
 
Substituting the respective values we get,
 
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}a & b+c & 1\\b & c+a & 1\\c & a+b & 1\end{vmatrix}$
 
Now expanding along the row $R_1$ we get
 
$\mid\bigtriangleup\mid=\frac{1}{2}\begin{bmatrix}a\begin{vmatrix}c+a & 1\\a+b & 1\end{vmatrix}-(b+c)\begin{vmatrix}b & 1\\c & 1\end{vmatrix}+1\begin{vmatrix}b & c+a\\c & a+b\end{vmatrix}\end{bmatrix}$
 
$=\frac{1}{2}[a[(c+a)-(a+b)]-(b+c)[b-c]+1[b(a+b)-c(c+a)]]$
 
$=\frac{1}{2}[a(c+a-a-b)-(b+c)(b-c)+1[ab+b^2-c^2-ac]]$
 
$=\frac{1}{2}[ac-ab-b^2+c^2+ab+b^2-c^2-ac]$
 
$=\frac{1}{2}[ac-ac]=0.$
Hence the area is 0,the points are collinear.

 

answered Feb 22, 2013 by sreemathi.v
edited Mar 6, 2013 by vijayalakshmi_ramakrishnans
 
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