Using the Distance Formula
Date: 05/03/97 at 16:31:30 From: ??? Subject: Finding a point on a line a certain distance from another point Hi Dr. Math, Can you help me with this problem? Given a line going through the points (10,10) and (20,15), find the coordinates of a third point on the line that is 3 units from the point (10,10). Thanks!
Date: 05/04/97 at 07:49:41 From: Doctor Anthony Subject: Re: Finding a point on a line a certain distance from another point To begin with, we need to find the equation of the line. The slope of the line is given by (difference of y's)/(difference of x's), which is equal to: (15-10)/(20-10) = 5/10 = 1/2 The equation of the line is: y-10 = (1/2)(x-10) y-10 = (1/2)x - 5 y = (1/2)x + 5 2y = x + 10 Let the point which is 3 units from (10,10) be (x,y). If a point is 3 units from another point, that means that the distance between the two points is three units. This in turn means that we can use the distance formula (which comes from the Pythagorean theorem). The distance formula says that the distance d between two points (x1,y1) and (x2,y2) is given by d^2 = (x2-x1)^2 + (y2-y1)^2. This makes a little more sense when you look at the following diagram: y-axis /|\ y2| /*(x2,y2) | / | | / | | / | y1| *----| | (x1,y1) | |__|____|_______\ x1 x2 / x-axis The distance between the two points (x1,y1) and (x2,y2) is the length of the hypotenuse of the above right triangle, right? We use the Pythagorean theorem to find the length of the hypotenuse of a right triangle. The length of the base of this triangle is (x2-x1). You can also write (x1-x2) because if this happens to be a negative number, it will be squared in the Pythagorean theorem. The height is (y2-y1), which again can also be written as (y1-y2). If we call the hypotenuse d, then from the Pythagorean theorem we can write: d^2 = (x2-x1)^2 + (y2-y1)^2 In our problem, d = 3 and we want to find the point (x,y) that is 3 units away from (10, 10). So we can write the distance formula as: (x-10)^2 + (y-10)^2 = 3^2 = 9. We substitute y = (1/2)x + 5 (the equation of our line) in for y: (x-10)^2 + (y-10)^2 = 9 Original distance formula (x-10)^2 + (x/2 + 5 - 10)^2 = 9 Substituting y = (1/2)x + 5 (x-10)^2 + (x/2 - 5)^2 = 9 Simplifying (x-10)^2 + (x/2 - 10/2)^2 = 9 Finding common denominator (x-10)^2 + ((x-10)/2)^2 = 9 Simplifying (x-10)^2 + (1/4)(x-10)^2 = 9 Factoring out (1/2)^2 = 1/4 (5/4)(x-10)^2 = 9 Factoring out (x-10)^2 (x-10)^2 = 36/5 x-10 = +- 6/sqrt(5) x = 10 +- 6/sqrt(5) x = 10 +- 6*sqrt(5)/5 We substitute x = 10 + 6*sqrt(5)/5 into the equation for our line: y = (1/2)x + 5 Equation for line = (1/2)(10 + 6*sqrt(5)/5) + 5 Substituting = 5 + 3sqrt(5)/5 + 5 Simplifying = 10 + 3sqrt(5)/5 The coordinates of the point that is 3 units from (10,10) are: [10+6sqrt(5)/5, 10+3qrt(5)/5] What if we use x = 10 - 6*sqrt(5)/5? Then we get another point that is 3 units away from (10,10), but in the opposite direction. Since your problem only asks for one point, we don't need to figure out the coordinates of the other point that is three units away from (10,10). Can you figure out why there are only two points on the line that are 3 units away from (10,10)? -Doctors Anthony and Rachel, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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