MHB How Do You Calculate the Slope to Intercept a Moving Point?

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To calculate the slope for a moving point, the standard slope formula (y2 - y1)/(x2 - x1) needs to be adapted for the moving point, referred to as point A. When point A is in motion, its position can be described using parametric equations based on its speed and direction, such as x(t) = t + c and y(t) = t + d, where c and d are the starting coordinates. The slope of the path can be determined by knowing the trajectory's slope and the initial coordinates. An example demonstrates this with a slope of 3/2 and a starting position of (3, 2), leading to specific equations for x(t) and y(t). Understanding these concepts allows for the calculation of the slope as the moving point approaches another point.
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Sorry for the very confusing title, I couldn't think of a different way to better explain my question shortly. Also please bare with me, I find this very hard to explain.

So I'm trying to find the slope between two points when one point is moving in a set direction with a set speed. Confused? I'm not the best with words.

Okay, so normally to find the slope of two points you use the simple formula (y2 - y1)/(x2 - x1). That's the quickest way to intersect a point (Now on going to be called point A) if point A is stationary. But what if point A is moving (In a set direction, not turning)?

Say point A is a someone you want to talk to, and they're walking. When walking to them you don't walk straight at them, you walk to a point where your two paths will cross at the same time. My question is how do you find the slope you decided to walk?

Hopefully that didn't just make things more complicated to understand. Any help will be appreciated!
 
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You could use parametric equations, if you have enough data.
Say the slope of the path of the moving object is 1. Then
x(t) = t + c and y(t) = t + d, where c and d are the starting
x and y coordinates, respectively. So you'd need to know
the slope of the trajectory of the moving object and the x and y
coordinates of its starting position.

Another example: Slope 3/2, (time, t, in seconds) and a starting position of (3, 2):

x(t) = 2t + 3, y(t) = 3t + 2.

For your problem you may then compute (y2 - y(t))/(x2 - x(t)).
 
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Thank you so much, it took me a couple read throughs but I understand it now!
 
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