If a straight line L perpendicular to the line 3 x-4 y=6 forms a triangle of area 6 square units with coordinate axes, then the minimum perpendicular distance from the point (1,1) to the line L is

Step-by-step solution

The line which is perpendicular to the given line will have a slope of - 4 3 . The equation of the line will look like 3 y+4 x=k . The line intersects the x -axis at ( k 4 , 0 ) an y -axis at (0, k 3 ) . The area of the triangle formed is given by aligned & = 1 2 k 3 k 4 6= 1 2 k^2 12 & k^2=12 12 k= 12 aligned Thus, the equation of line becomes 4 x+3 y= 12 or 4 x+3 y 12=0 For minimum distance from point (1,1) , we will take negative sign. So, the equation of line is 4 x+3 y-12=0 The required distance = | 4 1+3 1-12