Find position on line given start and end points

Question

I'm trying to solve this problem for myself which is for latitudes and longitudes. For the illustration below - I know ALL the variables EXCEPT $(a_1,b_1)$ which the latitude and longitude I need to calculate.

$X_o,Y_o$ - Circle Center

$X_2,Y_2$ - Outer perimeter

$M_d$ - Maximum distance I can go

$T_d$ - Total radius distance

How do I find the $(a_1,b_1)$ co-ords ?

Answer 1

Hint:-

1. Use section formula,

You can assume that the point $(x_2,y_2)$ divides the line joining $(x_1,y_1)$ and $(a_1,b_1)$ in the ratio $T_d$:$M_d$, externally.

2. Distance formula,

$\dfrac{a_1-x_1}{\cos\theta}=\dfrac{b_1-y_1}{\sin\theta}=M_D$ where $\theta$ is the slope of line joining $(x_1,y_1)$ and $(x_2,y_2.)$

Let us take, $(x_1,y_1)$ and $(x_2,y_2)$, $(0,0)$ and $(3,3)$ respectively and $M_d$ equal to $2\sqrt2$, for example. Here $\theta$ is $45^{\circ}$, using the formula, $(a_1,b_1)$ comes out to be $(2,2)$.

Answer 2

Let's imagine that $(a1,b1)$ is 70% of the way from $(x0,y0)$ to $(x2,y2)$. It's x- value will be 70% of the way from $x0$ to $x2$ and likewise it's y-value will be 70% of the way from $y0$ to $y2$. To find the coordinates of your point we can simply substitute 70% in my anecdote with $\dfrac{Md}{Td}$.

Hopefully the algebra here is not too confusing, but in your example this would give

\begin{align*} a1 &= \dfrac{Md}{Td}(x2-x0) + x0 \\[.5pc] b1 &= \dfrac{Md}{Td}(y2-y0) + y0 \end{align*}

For example

Imagine we have two points $A(2, 3)$ and $B(10, 12)$ and we want to find a point, C, that is 70% of the way from $A$ to $B$.

\begin{align*} x_C &= \dfrac{7}{10}(10-2) + 2 \\ &= 5.6 + 2 \\ &= 7.6 \\[.5pc] y_C &= \dfrac{7}{10}(12-3) + 3 \\ &= 6.3 + 3 \\ &= 9.3 \end{align*}

Point C would be at $(7.6, 9.3)$. Here's a quick Geogebra verification.

Although Geogebra is doing some rounding, you can tell that $AC/AB \approx 7/10$. In fact if you compute the distances exactly you will find that it is exactly $7/10$.