Given :The length of the perpendicular from the point (4, 1) on the line 3x - 4y + k = 0 is 2 units. Let P = (4, 1)Here x1=4,y1=1,a=3,b=−4 and d = 2 Now substitute x1=4 and y1=1 in the equation 3x−4y+k=0 ⇒|3.x1−4.y1+k|=|12−4+k|=|8+k| ⇒√a2+b2=√32+(−4)2=5 As we know that, the perpendicular distance d from P(x1,y1) to the line ax+by+c=0 is given by d=|
ax1+by1+c
√a2+b2
| ⇒d=|
3.x1−4.y1+k
√32+(−4)2
|=2 ⇒|8+k|=10 ⇒k=2 or -18 Hence, option C is the correct answer.