Number of triangles formed by the lines  $\mathrm{a}\left(\mathrm{x}+\mathrm{y}-2\right)+\mathrm{b}\left(3\mathrm{x}+4\mathrm{y}+7\right)=0$ and $5\mathrm{x}-4\mathrm{y}-1=0$

detailed solution

Correct option is D

Any two lines among three lines are parallel or if three line are passing through the same point then the number of triangles formed by those lines is zero, otherwise number of triangles is 1. The equation ax+y−2+b3x+4y−7=0 represents the set of lines passing through the point of intersection of x+y−2=0 and  3x+4y+7=0Eliminate  y4x+y−2−3x+4y−7=04x+4y−8−3x−4y+7=0x−1=0 Plug in x=1 in the equation  x+y−2=01+y−2=0y−1=0y=1 Hence the point of intersection of x+y−2=0 and 3x+4y+7=0  is 1,1 , but this point lies on the third line  5x−4y−1=0It implies that all three lines are concurrent, hence the number of triangles formed by the given three lines is zero