How to show wether 3 points lie on a straight line?

Question

A is at (1,1),B is at (3,4) and C is at (7,10).How to show ABC is a straight line? Thanx for helping.

Answer 1

do the point  slope form    from a to c

m = 3/2  (slope)

y-1=(3/2)(x-1)

y = (3/2)x-(1/2)

when you plug in B (3,4) they dont equal

4 not equal to (8/2)

Answer 2

take ax+by+c=0

its passes A,B and C in the above line and finding the values of a,b,c and plug the values of a,b,c in the equation of the straight line.

a+b+c=0

3a+4b+c=0

7a+10b+c=0

solving above equations

2a+3b=0 and 4a+6b=0

2a+3b=0 and 2a+3b=0 if u subtract these line u will get 0=0

hence the line satisfies

Answer 3

You could graph the points.  They have the same slope, but it's not clear they have the same intercepts.

Answer 4

There are several methods of doing that:-

# you can find the area of the triangle formed by them using coordinate geometry and if it would come 0, it is a straight line.

# you can find the slope of AB and slope of BC, if they are equal then the points are collinear.

# you can find the eqn. of AB by the concepts of straight lines and then get that eqn. satisfied by the third point C. if it is satisfied then it is a straight line

Answer 5

Use two of the points to derive the equation of the line, then show that the third point lies on the line.

slope m from A to B:

m = (4-1)/(3-1) = 3/2

point-slope equation of the line:

y-1 = (3/2)(x-1)

test equation with (7,10):

10-1 = (3/2)(7-1)

∴ ABC is a straight line.

http://www.flickr.com/photos/dwread/2787...

Answer 6

Find the gradients of AB and BC. If they are the same, then all 3 points lie on the same straight line.

Answer 7

Method 1 : Calculate the area of triangle formed by the these 3 points. If it is zero then these 3 points are collinear (area is 0 then there is no triangle, just a line). you can use the determinant to calculate the area of triangle.

Method 2 : Use vectors.

(good and easy)Method 3 : Form first equation of line using (1,1) & (3,4) and second equation of line by using (1,1) & (7,10). If the equations are same these three points lie on the same line.

Answer 8

Right, firstly, as you probably know, all points on a straight line (ie a line with no powers involved) will have their x and y coordinates in a fixed ratio. You also should know it is (x,y), so the line with A and B will have y as 1 when x is 1, and y as 4 when x is 3. Then you can find the differences of the x and y coordinates, so for every (3 - 1) x, or 2x, you go (4 - 1) y or 3y higher. So if you increase x by 4 (which is 2 times 2), you must therefore increase y by 6, so if you add translate it

(4)

(6) from B you will reach C, therefore proving it is on the line which includes the line segment AB.

Answer 9

Find the slope between each set:

AB, BC, and AC.

If they are all equal, they are collinear.

Slope = (y1 - y2) / (x1 - x2)

AB = (4 - 1) / (3 - 1) = 3 / 2

BC = (10 - 4) / (7 - 3) = 6 / 4 = 3 / 2

AC = (10 - 1) / (7 - 1) = 9 / 6 = 3 / 2

Answer 10

m AB = 3 / 2

m BC = 6 / 4 = 3 / 2

Line AB has same slope as line BC.

A , B and C on straight line.

Answer 11

From A to B, calculate ∆x and ∆y.  If you can multiply both of those by the same number to get the ∆x and∆y between A and C, then the 3 points are on the same line.

A to B, ∆x=2, ∆y=3.  A to C, ∆x=6, ∆y=9.  Multiplying the A-B ∆s by 3 will produce the A-C ∆s, so the points are all on the same line.

Answer 12

try graphing it in a cartesian plane. :D