How to decide the distance from a point to a line in your head

Question

Decide the distance from a point (1, -3) to a straight line (x, y) = (3 - t, 2 + 5t).

I know of a reliable process, but once I get to calculating the point from where the line is to meet the point (1, -3), the numbers get tricky to compute in your mind.

I first find the affine form of the straight line: 5x + y - 17 = 0 and since the basis is orthogonal, an orthogonal vector to the straight line is $v$ = (5, 1)

With it I can find an orthogonal straight line: -x + 5y + 16 = 0

And here is where it gets tricky for me. With the help of Gauss elimination, I manage to get the intersecting point to be P = (101/26, -63/26), which is for me quite tricky to do in my mind. Then I use the PB = (1, -3) - (101/26, -63/26) and calculate its length. All very time consuming and I cannot seem to get the correct answer! Do you know of any other method that is more suitable, or if I have done something wrong in the described method?

Answer 1

We need to minimize the distance$(d)$ $$d^2=(3-t-1)^2+(2+5t+3)^2$$ $$=26\left(t-\dfrac{23}{26}\right)^2+29-26\left(\dfrac{23}{26}\right)^2$$

$$\ge29-26\left(\dfrac{23}{26}\right)^2$$

the equality occurs if $t-\dfrac{23}{26}=0$

Answer 2

If you’re calculating this in your head, once you have the affine form of equation of the line, it’s much easier to use the standard distance formula $${\lvert ax_0+by_0+c\rvert\over\sqrt{a^2+b^2}}.$$

However, you can also use the method I described here to compute the distance directly from the parametric equation. This is easier than it looks at first because you can take advantage of the linearity of the determinant: $$\begin{vmatrix}\mathbf v \\ \mathbf v+\mathbf w\end{vmatrix} = \begin{vmatrix}\mathbf v\\\mathbf w\end{vmatrix}.$$ This means that you can take $\mathbf v$ as the difference of the two fixed points, here $(3,2)-(1,-3)=(2,5)$, and for $\mathbf w$ the direction vector $(-1,5)$ of the line. For your example, then, the final calculation ends up being $$\frac1{\sqrt{1^2+5^2}}\begin{vmatrix}2&5 \\ -1&5\end{vmatrix},$$ all of which can easily be done in one’s head.