logo

Ask Questions, Get Answers

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

Find the area of the triangle formed by the lines $y – x = \theta, x + y = \theta$ and $x – k = \theta$

$\begin {array} {1 1} (A)\;k\: sq\: units & \quad (B)\;2k\: sq\: units \\ (C)\;k^2\: sq\: units & \quad (D)\;2k^2\: sq\: units \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is $ \large\frac{1}{2}$$ | x_1 (y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
Equations of the given lines are :
$ \qquad y-x=0$---------(1)
$ \qquad x+y=0$----------(2)
$ \qquad x-k=0$----------(3)
Let us now find the points of intersection of the above lines
Solving equation (1) and (2) we get,
$\qquad x+y=0$
$ \qquad y-x=0$
We get $x=0, y=0$
Hence the point of intersection of line (1) and (2) is (0,0)
Next let us solve equation (2) and (3)
$ \qquad x+y=0$
$ \qquad x = k$
$\: \: \Rightarrow y = -k$
Hence the point of intersection of equation (2) and (3) is (k, -k)
Next let us solve equation (3) and (1)
$ \qquad x-k=0$
$ \qquad y-x=0$
$ \qquad y-k=0$
$ \Rightarrow y=k$ also $ x = k$
$ \therefore $ The point of intersection is (k, k)
Hence the vertices of the triangle formed by the three lines are (0,0) , (k, -k) and (k, k)
Now substituting the values in
area = $\large\frac{1}{2}$$ | x_1 (y_2-y_1)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ sq. units
area = $\large\frac{1}{2} | 0(-k \: -k)+k(k-0)+k(0+k)| $ square units
$ = \large\frac{1}{2}|k^2+k^2|$ sq units
$ = \large\frac{1}{2}$$ \times 2k^2$ sq. units or $k^2$ sq. units
answered May 20, 2014 by thanvigandhi_1
 

Related questions

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
...