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Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Q : 8     Find the area of the triangle formed by the lines  \small y-x=0,x+y=0  and    \small x-k=0

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Given equations of lines are
y-x=0 \ \ \ \ \ \ \ \ \ \ \ -(i)
x+y=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)
x-k=0 \ \ \ \ \ \ \ \ \ \ \ -(iii)
The point if intersection of (i)  and (ii) is  (0,0)
The point if intersection of (ii)  and (iii) is  (k,-k)
The point if intersection of (i)  and (iii) is  (k,k)
Therefore, the vertices of triangle formed by three lines are (0,0), (k,-k) \ and \ (k,k)
Now, we know that area of triangle whose vertices are (x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3)  is
A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|
A= \frac{1}{2}|0(-k-k)+k(k-0)+k(0+k)|
A= \frac{1}{2}|k^2+k^2|
A= \frac{1}{2}|2k^2|
A= k^2
Therefore, area of triangle  is  k^2 \ square \ units
 

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