Find all orthogonal vectors with the same length as a given vector

Find all vectors in $\mathbb{R}^2$ that have the same length as $A$ and are orthogonal to $A$ when $A$ is each of the following:

$A = (1,2)$ ;
$A = (1,-2)$ ;
$A = (2,-1)$ ;
$A = (-2,1)$ .

If $A = (1,2)$ , let $B = (b_1, b_2)$ be a vector such that $A \cdot B = 0$ and $\lVert A \rVert = \lVert B \rVert$ . Then we have
$A \cdot B = 0 \quad \implies \quad b_1 + 2b_2 = 0 \quad \implies \quad b_1 = -2b_2.$

Since the lengths are the same and $\lVert A \rVert = \sqrt{5}$ we also have

$\sqrt{5} = \lVert b \rVert = \sqrt{b_1^2 + b_2^2} = \sqrt{5b_2^2} \quad \implies \quad b_2 = \pm 1, \ b_1 = \mp 2.$

Therefore, $B = (2,-1)$ or $B = (-2,1)$ .
If $A = (1,-2)$ , let $B = (b_1, b_2)$ be a vector such that $A \cdot B = 0$ and $\lVert A \rVert = \lVert B \rVert$ . Then we have
$A \cdot B = 0 \quad \implies \quad b_1 - 2b_2 = 0 \quad \implies \quad b_1 = 2b_2.$

Since the lengths are the same and $\lVert A \rVert = \sqrt{5}$ we also have

$\sqrt{5} = \lVert b \rVert = \sqrt{b_1^2 + b_2^2} = \sqrt{5b_2^2} \quad \implies \quad b_2 = \pm 1, \ b_1 = \pm 2.$

Therefore, $B = (2,1)$ or $B = (-2,-1)$ .
If $A = (2,-1)$ , let $B = (b_1, b_2)$ be a vector such that $A \cdot B = 0$ and $\lVert A \rVert = \lVert B \rVert$ . Then we have
$A \cdot B = 0 \quad \implies \quad 2b_1 - b_2 = 0 \quad \implies \quad b_2 = 2b_1.$

Since the lengths are the same and $\lVert A \rVert = \sqrt{5}$ we also have

$\sqrt{5} = \lVert b \rVert = \sqrt{b_1^2 + b_2^2} = \sqrt{5b_1^2} \quad \implies \quad b_1 = \pm 1, \ b_2 = \pm 2.$

Therefore, $B = (1,2)$ or $B = (-1,-2)$ .
If $A = (-2,1)$ , let $B = (b_1, b_2)$ be a vector such that $A \cdot B = 0$ and $\lVert A \rVert = \lVert B \rVert$ . Then we have
$A \cdot B = 0 \quad \implies \quad -2b_1 + b_2 = 0 \quad \implies \quad b_2 = 2b_1.$

Since the lengths are the same and $\lVert A \rVert = \sqrt{5}$ we also have

$\sqrt{5} = \lVert b \rVert = \sqrt{b_1^2 + b_2^2} = \sqrt{5b_1^2} \quad \implies \quad b_1 = \pm 1, \ b_1 = \pm 2.$

Therefore, $B = (1,2)$ or $B = (-1,-2)$ .