Find a point on a line given, distance from startpoint

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To find the coordinates of point P on line L, which stretches from point A (x1, y1) to point B (x2, y2), a formula is derived using the slope of the line. The equation f(x) = y = (y2 - y1)/(x2 - x1)(x - x1) + y1 represents the line. To determine point P at a distance Z from A, the displacement is calculated as +/-Z along the slope. The angle of the slope is given by t = arctan[(y2 - y1)/(x2 - x1)]. The coordinates for point P can then be expressed as (x1 + (+/-Z)cos(t), f(x1 + (+/-Z)cos(t))).
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Hi,

I have a line L stretching from point A to ponit B. The startpoint A has the coordinates x1, y1 and endpoint B has the coordinates x2, y2.

I want a formula for calculating the coordinates for point P. Point P lies on the line L and with the distance Z from startpoint A.

BR Boston
 
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your equation is f(x)=y=(y2-y1)/(x2-x1)(x-x1) + y1 [this is just using y-y0 = m(x-x0)]

Now, a distance Z should be determined by a displacement +/-Z. There are two answers as written, if it is meant "in the direction of b" then replace +/-Z with just Z

Let t = arctan[(y2-y1)/(x2-x1)], this is the angle of the slope.

Then replace x in f(x) with x1+(+/-Z)cos(t). That is, you are finding y when x is at the point when you are +/-Zcos(t) from the coordinate x1. That is how much it changes when you move up or down the slope a displacement of +/-Z. Your point is (x1+(+/-Z)cos(t), f(x1+(+/-Z)cos(t))
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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