If the perpendicular distance between the point (1, 1) to the line 3x+4y+c=0 is 7, then the possible values of c are

Given $\perp r$ distance from (1,1) to 3x+4y+c=0 is '7'

$\begin{array}{l}\Rightarrow \frac{\left|3\left(1\right)+4\left(1\right)+c\right|}{\sqrt{3^2+4^2}}=7\\ \Rightarrow \frac{\left|c+7\right|}{5}=7\\ \Rightarrow \left|c+7\right|=35\\ \Rightarrow c+7=\pm 35\\ \therefore c+7=35c+7=-35\\ \Rightarrow c=28c=-42\end{array}$