Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
1485
4
avatar

do theese 3 points lie on the same lines?

(-4,-2) (2,2.5) (8,7)

or

(-10,4) (-3,2.8) (-17,6.8) pleeeease respond

 Dec 3, 2014

Best Answer 

 #3
avatar+26396 
+10

3 points lie on the same line, if the area of the triangle from 3 points is 0:

P1=(x1,y1)P2=(x2,y2)P3=(x3,y3) The area of a triangle with 3 Points is the determinant =|(x1x2)(x3x2)(y1y2)(y3y2)| area =(x1x2)(y3y2)(y1y2)(x3x2)

I

do theese 3 points lie on the same lines?

(-4,-2) (2,2.5) (8,7)

\\P_1(x_1 = -4,\ y_1 = -2 ) \\ P_2(x_2 = 2,\ y_2 = 2.5 ) \\ P_3(x_3 = 8,\ y_3 = 7 ) \\ \text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} \\ \text{ area } = (-4-2) *(7-2.5) - ( (-2)-2.5)*(8-2)} \\ \text{ area } = (-6) *(4.5) - ( -4.5)*(6)} \\ \text{ area } = (-27) - ( -27)} \\ \text{ area } = -27 + 27 = 0 }

area = 0: The 3 Points lie on the same line.

II

do theese 3 points lie on the same lines?

(-10,4) (-3,2.8) (-17,6.8)

\\P_1(x_1 = -10,\ y_1 = 4 ) \\ P_2(x_2 = -3,\ y_2 = 2.8 ) \\ P_3(x_3 = -17,\ y_3 = 6.8 ) \\ \text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} \\ \text{ area } = (-10-(-3)) *(6.8-2.8) - ( 4-2.8)*(-17-(-3))} \\ \text{ area } = (-10+3) *(4) - ( 1.2)*(-17+3)} \\ \text{ area } = (-7) *(4) - ( 1.2)*(-14)} \\ \text{ area } = (-28) - ( -16.8 )} \\ \text{ area } = -28 + 16.8 = - 11.2 \ne 0 }

The area is not 0. The 3 Points lie not on the same line.

 Dec 3, 2014
 #1
avatar+23254 
+8

Whenever you have a straight line, it will have the same slope no matter what points you use.

To find whether or not these points are on one straight line:  (-4,-2)   (2,2.5)   (8,7)

Find the slope using the first two points; then find the slope using the last two points; if these slopes are the same, then the three points are on the same line. If they aren't the same, then they don't lie on the same line.

Formula for slope:  m  =  (y2 - y1) / (x2 - x1)

Can you take it from here?

 Dec 3, 2014
 #2
avatar
0

oh my gosh! yes ! thank you so much! because of that i will probably move on to 9th grade algebra 2! thanks again!

 Dec 3, 2014
 #3
avatar+26396 
+10
Best Answer

3 points lie on the same line, if the area of the triangle from 3 points is 0:

P1=(x1,y1)P2=(x2,y2)P3=(x3,y3) The area of a triangle with 3 Points is the determinant =|(x1x2)(x3x2)(y1y2)(y3y2)| area =(x1x2)(y3y2)(y1y2)(x3x2)

I

do theese 3 points lie on the same lines?

(-4,-2) (2,2.5) (8,7)

\\P_1(x_1 = -4,\ y_1 = -2 ) \\ P_2(x_2 = 2,\ y_2 = 2.5 ) \\ P_3(x_3 = 8,\ y_3 = 7 ) \\ \text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} \\ \text{ area } = (-4-2) *(7-2.5) - ( (-2)-2.5)*(8-2)} \\ \text{ area } = (-6) *(4.5) - ( -4.5)*(6)} \\ \text{ area } = (-27) - ( -27)} \\ \text{ area } = -27 + 27 = 0 }

area = 0: The 3 Points lie on the same line.

II

do theese 3 points lie on the same lines?

(-10,4) (-3,2.8) (-17,6.8)

\\P_1(x_1 = -10,\ y_1 = 4 ) \\ P_2(x_2 = -3,\ y_2 = 2.8 ) \\ P_3(x_3 = -17,\ y_3 = 6.8 ) \\ \text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} \\ \text{ area } = (-10-(-3)) *(6.8-2.8) - ( 4-2.8)*(-17-(-3))} \\ \text{ area } = (-10+3) *(4) - ( 1.2)*(-17+3)} \\ \text{ area } = (-7) *(4) - ( 1.2)*(-14)} \\ \text{ area } = (-28) - ( -16.8 )} \\ \text{ area } = -28 + 16.8 = - 11.2 \ne 0 }

The area is not 0. The 3 Points lie not on the same line.

heureka Dec 3, 2014
 #4
avatar+118703 
0

That is a rather unique way of looking at this problem Heureka.  

Thanks Geno and Heureka   

 Dec 3, 2014

1 Online Users