How do you find these points (-2,-1) (1,5) (3,10) lie on the same line without plotting the points on a graph?
2007-07-08 14:04:48 · 10 answers · asked by Okay.... 3 in Science & Mathematics ➔ Mathematics
Find the slope from point A to point B. Then find the slope from point B to point C. If the two slopes are equal then the points are all on the same line.
in this case slope from A(-2, -1) and B(1,5) is equal to:
change in y / change in x
= (-1 -5) / (-2 -1) = -6/-3 = 2
slope from B(1,5) to c(3,10) is equal to:
change in y / change in x
= (5-10)/(1-3) = -5/-2 = 2.5
since the slope are different the three points do not lie on the same straight line.
2007-07-08 14:07:48 · answer #1 · answered by ignoramus_the_great 7 · 0⤊ 0⤋
Find the slope between each point. If they match, they lie on the same line.
m=(y2-y1)/(x2-x1)
(1,5)(-2,-1):
(5--1)/(1--2)=6/3=2
(3,10)(-2,-1):
(10--1)/(3--2)=11/5
(1,5)(3,10):
(10-5)/(3-1)=5/2
So, the slopes don't match, therefore, they aren't the same line.
2007-07-08 21:11:48 · answer #2 · answered by Shauna 3 · 0⤊ 0⤋
if they all have the same slope,
try (-2, -1) and (1,5), then (1,5) and (3,10) then (-2,-1) and (3,10) the process is this
y1 -y2 / x1-x2
plug in for all the above mentioned if they all equal the same then they are all lined in
2007-07-08 21:13:17 · answer #3 · answered by Leo 3 · 0⤊ 0⤋
Find the slope of two different pairs of points. (With three points, one of them will be use in both slope calculations.)
If both slopes are the same, and if both slopes have one enpoint in common, then all three points are on the same line.
m1 = (-1-5)/(-2-1) = -6/-3 = 2
m2 = (-1-10)/(-2-3)= -11/-5 = 2.2
They don't have the same slope, so they can't be colinear.
2007-07-08 21:13:32 · answer #4 · answered by Tony The Dad 3 · 0⤊ 0⤋
you find the slope of two points at a time with fomula of (change in y / change in x)
after you find the slope of first and second point. And second and third. if the slope is the same. then they lie on the same line. but in this case I don't think it does.
2007-07-08 21:11:16 · answer #5 · answered by jkim972 3 · 0⤊ 0⤋
Can 3 points exist in a line on a plane in space? Does the line have to be straight? I think a curved line, is still defined as a line. :)
2007-07-08 21:10:20 · answer #6 · answered by Narnia 2 · 0⤊ 0⤋
you can find the equation of the line using 2 of the points. then substitute the co-ordinates of the 3rd point and show that it satisfies the equation you found. Thus it must lie on that line too.
If it does not satisfy it then it will not be on the same line.
In fact, they are not on the same line.
2007-07-08 21:10:44 · answer #7 · answered by swd 6 · 0⤊ 0⤋
y-subAB = (5 - -1)/(1 - -2) = 6/3 = 2
y-subBC = (10 - 5)/(3 - 1) = 5/2 = 2.5
These segments have different slopes, so they can't be colinear.
2007-07-08 21:12:00 · answer #8 · answered by TychaBrahe 7 · 0⤊ 0⤋
Use A TI-83, 84 or higher graphing calculator. Sorry I do not believe there are any tricks.
2007-07-08 21:07:39 · answer #9 · answered by Youngboss 3 · 0⤊ 1⤋
I dont think those coordinates are on the same line..
slopes are all different.
2007-07-08 21:10:17 · answer #10 · answered by MathMaelstrom 1 · 0⤊ 0⤋