Calculating a point on a straight line which is "t" distance away from a fixed point on the same straight line using parametric equations

Question

I found only one question relevant in MSe but the answer is not helpful for my problem.

Question : I have a line $(L_{1})$ $y = 3 \cdot x+2$. I know a point on this line $X_{0} = (0,2)$. I want to find another point on this line which is $t = 2$ units away from $X_{0}$.

Issue: I am able to find a point on this line but it is not $t = 2$ units away from the point $X_{0}$. I want to use only parametric equation to solve this problem.

My solution: If I have a straight line of the form $a^{T} \cdot X + b = 0$ then the parametric form is $X(t) = X(0) + t \cdot R \cdot \frac{a}{\|a\|} \forall t \in (\inf, -\inf)$.

Where $t$ = distance between $X(0)$ and $X(t)$. and $R = \begin{bmatrix}0 & -1\\1 & 0 \end{bmatrix}$.

Calculation: $y = 3 \cdot x+2$ can be written as $\begin{bmatrix} 3 \\ -1 \end{bmatrix} \cdot X + 2$. Where $X = \begin{bmatrix} x \\y \end{bmatrix}$. Then I found $a = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$ by comparing with the formula I have above. $X(0) = \begin{bmatrix} 0 \\ 2 \end{bmatrix}$ and $t = 2$. By substituting $X(0), t, a$ in the formula of $X(t)$, I found $X(t = 2) = \begin{bmatrix} 0.6324 \\ 3.8973 \end{bmatrix}$.

Problem: when I substitute $X(t=2)$ in $L_{1}$, I find that the value of $L_{1} = 0$. So the point is on the line $L_{1}$ BUT when I find the distance between $X(t=2) \ and \ X(0)$, it is not equal to $2$ units!! :(. Can someone help me find where my mistake is? Am I dealing with change of basis vectors or something? I ran out of ideas.

Edit: The solution I did above is correct. Thanks for the answer below I was able to find that the distance between my $X(t=2) \ and \ X(0)$ is 2 units. It was a genuine calculation mistake.

Answer 1

Your mistake must lie somewhere in your final computation of the distance between $X(2)$ and $X(0)$. With the rounded off values that you gave in your question, the computed distance is within a few hundred thousandths of $2$, and if you use exact values for the coordinates of $X(2)$, the distance is exactly $2$.

Answer 2

Find the co-ordinates satisfying: $$y=3x+2$$ $$\sqrt{(x-0)^2+(y-2)^2}=2$$ That is to say, solve: $$x^2+(3x)^2=4$$ Which gives us $$x=\pm\frac{\sqrt{10}}{5}\to y=\frac{10\pm3\sqrt{10}}{5}$$

What I've used is the equation of the existing line and the equation for the length of a straight line.

Here they are both graphed

Answer 3

In vector form, the line $y = 3x+2$ can be expressed as $P_0(t) = (0,2)+t(1,3)$. The direction of this line is the direction of the vector $(1,3)$. We can make this a unit vector by dividing by $\|(1,3)\| = \sqrt{10}$.

Then we get $P(t) = (0,2)+\dfrac{t}{\sqrt{10}}(1,3)$ and the parameter, $t$ becomes the directed distance on the line from the point $(0,2)$.

There are two points on the line that are a distance of 2 from the point $(0,2)$

$$P(2) = \left(\sqrt{\frac25}, 2 + 3\sqrt{\frac 25} \right) = (0.63246, 3.8974)$$ $$P(-2) = \left(-\sqrt{\frac25}, 2 - 3\sqrt{\frac 25} \right) =(-0.63246, 0.10263) $$