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collinear
- Thread starter elledo
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D
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Please follow the rules of posting in this forum, as enunciated at:im . taking geometry but this is more on the algebra side, so i think it fits here
given A (7,-1) and B(1,1) find coordinates of a point c thats collinear to A and B
(must solve algebraically)
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Please share your work/thoughts about this assignment.
Hint: The equation of a line passing through (x1,y1) and x2,y2) is:
\(\displaystyle \frac{y \ - \ y_1}{x \ - \ x_1} \ = \ \frac{y_2 \ - \ y_1}{x_2 \ - \ x_1} \ \)
If you are still stuck - tell us exactly where you are getting lost.
HallsofIvy
Elite Member
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- Jan 27, 2012
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- 7,763
First, do you understand that there are an infinite number of points on a line so an infinite number of correct answers to this problem?
The simplest point to find is the one exactly half way between (7, -1) and (8, 1) by averaging the coordinates. Half way between 7 and 8 is (7+ 8)/2= 7.5 and half way between -1 and 1 is (-1+ 1)/2= 0. (7.5, 0) is on the line, exactly half way between (7. -1) and (8, 1).
More generally, you can find any point on the line between two given points by first finding the equation of the line through the two points. The equation of any (non-vertical) line in an xy-coordinate system can be written y= ax+ b for a and b some numbers. Since (7, -1) is on the line -1= 7a+ b. Since (8, 1) is on the line, 1= 8a+ b. Subtracting the first equation from the second, 2= a. So we have -1= 7(2)+ b or b= -15. Any (x, y) point on this line must satisfy y= 2x- 15. Notice that x= 7.5 gives y= 2(7.5)- 15= 0 so, again, (7.5, 0). Or, for the easiest computation, taking x= 0, y= 2(0)- 15 so (0, -15) is also a point on the line. Taking x to be any value at all gives us infinitely many points on the line. If you want to impress (or perhaps irritate) your teacher, take \(\displaystyle x= \pi\) to get the point \(\displaystyle (\pi, 2\pi- 15)\).
The simplest point to find is the one exactly half way between (7, -1) and (8, 1) by averaging the coordinates. Half way between 7 and 8 is (7+ 8)/2= 7.5 and half way between -1 and 1 is (-1+ 1)/2= 0. (7.5, 0) is on the line, exactly half way between (7. -1) and (8, 1).
More generally, you can find any point on the line between two given points by first finding the equation of the line through the two points. The equation of any (non-vertical) line in an xy-coordinate system can be written y= ax+ b for a and b some numbers. Since (7, -1) is on the line -1= 7a+ b. Since (8, 1) is on the line, 1= 8a+ b. Subtracting the first equation from the second, 2= a. So we have -1= 7(2)+ b or b= -15. Any (x, y) point on this line must satisfy y= 2x- 15. Notice that x= 7.5 gives y= 2(7.5)- 15= 0 so, again, (7.5, 0). Or, for the easiest computation, taking x= 0, y= 2(0)- 15 so (0, -15) is also a point on the line. Taking x to be any value at all gives us infinitely many points on the line. If you want to impress (or perhaps irritate) your teacher, take \(\displaystyle x= \pi\) to get the point \(\displaystyle (\pi, 2\pi- 15)\).