Mobile robot 1

Define the vector from A to B:

$\large{\vec{v_{AB}} = (13 - 3) \hat{\imath} + (9 - 0) \hat{\jmath} = 10 \hat{\imath} + 9 \hat{\jmath}}$

The first unit vector of a pair of orthogonal unit vectors that will define the new coordinate system:

$\large{\vec{u_1} = \frac{10 \hat{\imath} + 9 \hat{\jmath}}{\sqrt{181}}}$

And the second:

$\large{\vec{u_2} = \frac{-9 \hat{\imath} + 10 \hat{\jmath}}{\sqrt{181}}}$

The displacement of the small object from the origin in the original coordinate system is:

$\large{\vec{D}_x} = 13 + 6 \vec{{u}_1}_x + 5 \vec{{u}_2}_x = 13 + \frac{60}{\sqrt{181}} - \frac{45}{\sqrt{181}} \approx \boxed{14.115} \\ \large{\vec{D}_y} = 9 + 6 \vec{{u}_1}_y + 5 \vec{{u}_2}_y = 9 + \frac{54}{\sqrt{181}} + \frac{50}{\sqrt{181}} \approx \boxed{16.730}$

You may solve this by using rotation and translation $[4 \times 4 ]$ homogenous matrices.

As shown in the above figure, there two Transformations $T_1 \ \text{and} \ \ T_2$ , where

$T_1 =$ translation matrix from point $(0,0)$ to point $(3,0) = \left(\begin{array}{ccc} 1 & 0& 0 & \color{#D61F06} 3\\ 0 &1 &0 & \color{#D61F06} 0 \\ 0 & 0 &1 &0 \\ 0 & 0 &0 &1 \\ \end{array} \right)$

$T_2 =$ translation matrix from point $(3,0)$ to point $(13,9)$ followed by rotation of $\angle \theta = \left(\begin{array}{ccc} \cos(\color{#624F41}\theta) & -\sin(\color{#624F41}\theta) &0 & \color{#D61F06}10 \\ \sin(\color{#624F41}\theta) & \cos(\color{#624F41}\theta) & 0 & \color{#D61F06}9\\ 0 &0 &1 &0 \\ 0 & 0 &0 &1 \\ \end{array} \right)$

$\begin{aligned} \angle \color{#624F41}\theta & = \text{ rontation angle about } \ \ z-axis \ \ \text{by} \ \ \angle \tan^{-1}(\frac{9}{10})\ \text{and} \\ &\sin({\theta }) = \dfrac{9}{\sqrt{181}} \ \ \text{\&} \ \ \cos({\theta }) = \dfrac{10}{\sqrt{181}} \end{aligned}$

Then the total transformation $T = T_1 \ T_2 = \left(\begin{array}{ccc} \dfrac{10}{\sqrt{181}} & -\dfrac{9}{\sqrt{181}} &0 & \color{#D61F06}13 \\ \dfrac{9}{\sqrt{181}} & \dfrac{10}{\sqrt{181}} & 0 & \color{#D61F06}9\\ 0 &0 &1 &0 \\ 0 & 0 &0 &1 \\ \end{array} \right) \\$

Any point expressed in frame $(x_1,y_1)$ can be expressed in frame $(x_0,y_0)$ by using the transformation matrix $T$

Point $C$ in frame $(x_0,y_0) = T$ [ $C$ expressed in frame $(x_1,y_1) ] = T \left(\begin{array}{c} &6 \\ &5\\ &0 \\ &1 \\ \end{array} \right) = \left(\begin{array}{c} &\frac{60}{\sqrt{181}} - \frac{45}{\sqrt{181} } + 13 \\ &\frac{54}{\sqrt{181}} +\frac{50}{\sqrt{181}} + 9 \\ &0 \\ &1 \\ \end{array} \right) = \left(\begin{array}{c} &14.115 \\ & 16.730\\ &0 \\ &1 \\ \end{array} \right)$