Triangle

Here's one way to do this...but maybe not the most efficient

Move  "B"  to the origin   ....so   ( 1 - 1, - 4 + 4)  = (0, 0)

Apply the same transformation to "A"  and we have   

(5 -1, 3 + 4)  =  ( 4, 7)

Rotating this point 90° counter-clockwise produces (-7, 4)

Now.......reverse the original transformation  and we have

(-7 + 1, 4 - 4)   =    (-6, 0)  = A'

A triangle with vertices A(5,3), B(1,-4) and C(-3,1) is rotated 90 degrees counterclockwise about vertex B.

Find the co-ordinates of the image of vertex A.

\(\begin{array}{llcll} \text{Formula Rotation:} & \boxed{\vec{A}' = (\vec{A}-\vec{B_{rotation\ axis}})\cdot D +\vec{B_{rotation\ axis}} } \\ & \vec{A} \text{ before rotation} \\ & \vec{A}' \text{ after rotation} \\ & \text{$\vec{B_{rotation\ axis}}$ at $\binom{1}{-4} $ } \\ & \overset{\curvearrowleft}{D}_{\varphi} = \begin{pmatrix} \cos(\varphi) & \sin(\varphi) \\ -\sin(\varphi) & \cos(\varphi) \end{pmatrix} \ \text{ Matrix of rotation counterclockwise}\\ & \overset{\curvearrowleft}{D}_{90^{\circ}} =\begin{pmatrix} \cos(90^{\circ}) & \sin(90^{\circ}) \\-\sin(90^{\circ}) & -\cos(90^{\circ}) \end{pmatrix} =\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \\ \end{array} \)

\(\begin{array}{llcll} \text{Formula Rotation:} & \boxed{\vec{A}' = (\vec{A}-\vec{B_{rotation\ axis}})\cdot D +\vec{B_{rotation\ axis}} } \\ & \begin{array}{|rcll|} \hline \vec{A}' &=& \left(\dbinom{5}{3}-\dbinom{1}{-4} \right)\cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} +\dbinom{1}{-4} \\ &=& \dbinom{4}{7} \cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} +\dbinom{1}{-4} \\ &=& \dbinom{-7}{4}+\dbinom{1}{-4} \\ &=& \dbinom{-6}{0} \\ \hline \end{array} \end{array} \)