logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Determinants
0 votes

Find area of the triangle with vertices at the point given in (1, 0), (6, 0), (4, 3)

$\begin{array}{1 1} 15/2 \\ 30 \\ 15 \\ 60 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
  • $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
  • $\bigtriangleup=\frac{1}{2}[x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2y_3-y_2x_3)]$
Let $(x_1,y_1)$ be=(1,0)
 
Let $(x_2,y_2)$ be=(6,0)
 
Let $(x_3,y_3)$ be=(4,3)
 
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,
 
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}1 & 0 & 1\\6 & 0 & 1\\4 & 3 & 1\end{vmatrix}$
 
Now expanding along the row $R_1$ we get
 
$\frac{1}{2}\begin{bmatrix}1\begin{vmatrix}0 & 1\\3 & 1\end{vmatrix}-0\begin{vmatrix}6 & 1\\4 & 1\end{vmatrix}+1\begin{vmatrix}6 & 0\\4 & 3\end{vmatrix}\end{bmatrix}$
 
$=\frac{1}{2}[1(-3)-0+1(18)]$
 
$=\frac{1}{2}[-3+18]=15/2.$
 
Hence the area is 15/2sq.units.

 

answered Feb 22, 2013 by sreemathi.v
edited Feb 24, 2013 by vijayalakshmi_ramakrishnans
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...