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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

Guest Dec 3, 2014

Best Answer 

 #3
avatar+19620 
+10

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

$$\\ P_1 = (x_1,y_1) \\
P_2 = (x_2,y_2) \\
P_3 = (x_3,y_3) \\
\small{\text{ The area of a triangle with 3 Points is the determinant } = \left| \begin{array}{cc}
(x_1-x_2) & (x_3-x_2) \\
(y_1-y_2) & (y_3-y_2) \\
\end{array} \right|\\\\

\boxed{ \text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} }$$

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
 #1
avatar+17744 
+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?

geno3141  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!

Guest Dec 3, 2014
 #3
avatar+19620 
+10
Best Answer

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

$$\\ P_1 = (x_1,y_1) \\
P_2 = (x_2,y_2) \\
P_3 = (x_3,y_3) \\
\small{\text{ The area of a triangle with 3 Points is the determinant } = \left| \begin{array}{cc}
(x_1-x_2) & (x_3-x_2) \\
(y_1-y_2) & (y_3-y_2) \\
\end{array} \right|\\\\

\boxed{ \text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} }$$

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+92775 
0

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

Thanks Geno and Heureka   

Melody  Dec 3, 2014

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