One vertex of a square $ABCD$ is $A(-1, 1)$ and the equation of one diagonal $BD$ is $3x+y-8=0$ then $C$=

Let point $C (x,y)$.

Slope of $AC = \dfrac{{y - 1}}{{x + 1}}$

and slope of $BD = 3x + y - 8 = 0$

$\begin{array}{l} y=-3x+8\to \left( i \right) \\ \Rightarrow Slope\, \, of\, \, BD\, \, is\, -3 \\ \Rightarrow AC\bot BD \\ \Rightarrow \dfrac { { y-1 } }{ { x+1 } } \times -3=-1 \\ \Rightarrow 3\left( { y-1 } \right) =x+1 \\ \Rightarrow 3y-3=x+1 \\ \Rightarrow 3y-x=4\to \left( { ii } \right) \end{array}$

Solving equation (i) and (ii)

$\begin{array}{l} \Rightarrow \dfrac { x }{ { 20 } } =\dfrac { y }{ { 20 } } =\dfrac { 1 }{ { 20 } } \\ \therefore x=2,y=2 \\ \Rightarrow then,\, \, by\, \, { { using } }\, \, mid\, formula,\dfrac { { x-1 } }{ 2 } =2,x=5 \\ \dfrac { { y+1 } }{ 2 } =2,y=3 \end{array}$