Given let (1,1)=(x1,y1) Let d1 is the distance of the line 3x+4y=1 from the point (1,1) . Let d2 is the distance of the line 4x+3y+2k=0 from the point (1,1) . The two straight lines 3x + 4y = 1 and 4x + 3y + 2k = 0 are equidistant from the point (1, 1). Therefore, d1=d2 The distance of the line ax+by+c=0 from the point (x1,y1) is given by ,d=
|ax1+by1+c|
√a2+b2
Hence, the distance of the line 3x + 4y - 1 = 0 from the point (1, 1) is d1=
|(3)(1)+(4)(1)−1|
√32+42
d1=
6
5
Similarly, the distance of the line 4x + 3y + 2k = 0 from the point (1, 1) is d2=