As an alternative approach, you can look for the vector perpendicular to the direction of the line (i.e. $\langle2,10\rangle$) that passes through the point $P=(2,6)$.

The points on the line are $Q_t=\langle1,-1\rangle + t\langle 2,10\rangle$. We just need to find the one for which the vector $\vec{PQ_t}$ joining it to $P$ is perpendicular to $\langle2,10\rangle$.

$$\vec{PQ_t}=\vec{OQ_t} - \vec{OP} = \langle1,-1\rangle + t\langle 2,10\rangle - \langle 2,6 \rangle = \langle 2t-1, 10t-7 \rangle$$

For this to be perpendicular to $\langle2,10\rangle$, we require that the scalar (dot) product of $\langle2,10\rangle$ and $\vec{PQ_t}$ be equal to zero.

That is \begin{align}\vec{PQ_t} \cdot \langle2,10\rangle = \langle 2t-1, 10t-7 \rangle \cdot \langle2,10\rangle &= 0 \\ 2(2t-1) + 10(10t-7) &= 0 \\ 104t - 72 &= 0 \\ t &= \tfrac9{13} \end{align}

Thus the vector $\vec{PQ_t}=\langle 2t-1, 10t-7 \rangle$ with $t=\frac9{13}$ is the perpendicular that joins the line and the point $P$. It's length is the perpendicular distance between the line and $P$.