Let $l$={ $t\left( \begin{matrix} 3\\ 4\end{matrix} \right):t\in\mathbb{R}$ }. Find the distance between $l$ and $\left( \begin{matrix} 1\\ 2\end{matrix} \right)$.
Solution. Note that $l$={ $\left( \begin{matrix} x\\ y\end{matrix} \right): -4x+3y=0$ }.
So, the distance is $\dfrac {\left| -4.1+3.2\right| } {\sqrt {\left( -4\right) ^{2}+3^{2}}}=2/5$.
My question is: Why $l$={ $\left( \begin{matrix} x\\ y\end{matrix} \right): -4x+3y=0$ }? How did we obtain this?
On
Assuming you know single-variable calculus, you can think of this problem as minimzing the distance between $l$ and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
$$ \text{dist}(u,v) = \left \| u-v \right \| = \sqrt{\left \langle u-v,u-v \right \rangle} $$
To find a function of $t$ giving the distance between $l$ and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ we just use the distance formula.
$$ d(t) = \sqrt{(3t-1)^2 + (4t-2)^2} = \sqrt{25t^2 - 22t + 5} $$
Now to find the value of $t$ that minimizes $d(t)$ we just solve $d'(t)=0$.
This gives us a value of $t=\frac{11}{25}$
now evaluating $d \left ( \frac{11}{25} \right )$ we arrive at the answer of $2/5$
In fact this is how that expression you gave for the distance between a line and a point is derived.
Hint: eliminate $t$ between $x=3 t$ and $y = 4t$.