Number of triangle formed by the lines $x-y+3=0,2 x-y+3=0,3 x-y+2=0$ and $x+y-3=0$ is
Number of triangle formed by the lines $x-y+3=0,2 x-y+3=0,3 x-y+2=0$ and $x+y-3=0$ is
4
6
3
2
Solution
Slope of line $x-y+3=0$ is 1 .
Slope of line $2 x-y+3=0$ is 2 .
Slope of line $3 x-y+2=0$ is 3 .
Slope of line $x+y-3=0$ is -1 .
As the slope of lines are different, there are 4 concurrent lines.
Number of triangles formed by these lines
$
={ }^4 C_3=4
$