Question

Determine whether the following points (-2,-2),(-6,-2) and(-2,2) are collinear or not.

Answer 1

Answer:

The points (– 2, – 2), (– 6, – 2) and (– 2, 2) are not collinear.

$\qquad$_____________

Step-by-step explanation:

Let the points be named as :

• A = (– 2, – 2)
• B = (– 6, – 2)
• C = (– 2, 2)

$\:$

Now, let's find the distance between of :

• AB
• BC
• AC

$\:$

Firstly, the distance of AB :

$\rm{ : \implies \: \sqrt{ (x_2 - x_1) ^{2} + {( y_2 - y_1) }^{2} } }$

$\rm{ \: \implies \: \sqrt{(-6+2)^2+(-2-2)^2} }$

$\rm{ : \implies \: \sqrt{ {(-4) }^{2} - {( -4)}^{2} } }$

$\rm:\implies~ \sqrt{16+16}$

$\pmb{\rm{:\implies~ AB = 4\sqrt{2}~units}}$

Now, the distance of BC :

$\rm:\implies~ \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\rm:\implies~ \sqrt{(-2+6)^2+(2-2)^2}$

$\rm:\implies~ \sqrt{(4)^2}$

$\pmb{\rm{:\implies~ BC =4~ units}}$

$\:$

Finally, the distance of AC :

$\rm:\implies~ \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\rm:\implies~ \sqrt{(-2-2)^2+(2-2)^2}$

$\rm:\implies~ \sqrt{(-4)^2}$

$\pmb{\rm{:\implies~ AC=4~units}}$

$\:$

We know that either of the three conditions have to be satisfied for the points in order to be collinear :

• AC + BC = AB
• AB + BC = AC
• AB + AC = BC

$\:$

Let's check for AC + BC = AB firstly :

$\rm:\implies~ AC +BC+AB$

$\rm:\implies~ 4~ units+4~units=4\sqrt{2}~units$

$\rm:\implies~ 8~units\neq 4\sqrt{2}~units$

So, AC + BC ≠ AB.

$\:$

Now, for AB + BC = AC :

$\rm:\implies~ AB +BC = AC$

$\rm:\implies~ 4\sqrt{2}~units+4~units=4~units$

$\rm:\implies~ 4\sqrt{2}~units+4~units\neq4~units$

So, AB + BC ≠ AC.

$\:$

Finally, for AB + AC = BC :

$\rm:\implies~ AB +AC=BC$

$\rm:\implies~ 4\sqrt{2}~units+4~units=4~units$

$\rm:\implies~ 4\sqrt{2}~units+4~units\neq4~units$

So, AB + AC ≠ BC.

$\qquad$_____________

None of the conditions have been satisfied.

Therefore, the points (– 2, – 2), (– 6, – 2) and (– 2, 2) aren't collinear.

Answer 2

Answer:

The points (– 2, – 2), (– 6, – 2) and (– 2, 2) are not collinear.

Let the points be named as :

A = (– 2, – 2)

B = (– 6, – 2)

C = (– 2, 2)  

Now, let's find the distance between of :  

AB

BC

AC  

Firstly, the distance of AB :

Now, the distance of BC :

Finally, the distance of AC :

We know that either of the three conditions have to be satisfied for the points in order to be collinear :

AC + BC = AB

AB + BC = AC

AB + AC = BC

Let's check for AC + BC = AB firstly :

So, AC + BC ≠ AB.

Now, for AB + BC = AC :

So, AB + BC ≠ AC.

Finally, for AB + AC = BC :

So, AB + AC ≠ BC.

None of the conditions have been satisfied.

Therefore, the points (– 2, – 2), (– 6, – 2) and (– 2, 2) aren't collinear.

Step-by-step explanation:

thanks..