Question

(–2, –8), (2,–3) (6,2) Determine whether the given set of points in each case are collinear or not.

Answer 1

1st distance between points A(-2,-8).B(2,-3)
By distance formula
$\sqrt{( 2 - ( - 2) {}^{2} +( - 3 - ( - 8) {}^{2} } \\ \sqrt{( {4}^{2} ) + ( - 5) {}^{2} } \\ \sqrt{16 + 25} \\ \sqrt{41} .....1$
2nd distance between points B (2,-3),C(6,2)

$\sqrt{ {(6 - 2)}^{2} + {(2 - ( - 3)}^{2} } \\ \sqrt{ {4}^{2} + {5}^{2} } \\ \sqrt{41} ....2$
3rd distance between points A(-2,-8),C(6,2)
$\sqrt{ {(6 - ( - 2)}^{2} + (2 - ( - 8) {}^{2} } \\ \sqrt{ {8}^{2} + {10}^{2} } \\ \sqrt{64 + 100} \\ \sqrt{164} \\ \sqrt{4 \times 41} \\ 2 \sqrt{41} .....3$
from 1,2,3
$\sqrt{41} + \sqrt{41} = 2 \sqrt{41} \\ 2 \sqrt{41} = 2 \sqrt{41}$
AB+BC=AC
Therefore, the given points are collinear.
Hope this helps you :-))))
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Answer 2

Hi,

***************************************

We know that ,

When the area of a triangle ∆ABC

is zero

then the three points A , B and C

said to be collinear points .
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Here ,

Let A(-2,-8) = ( x1,y1)

B(2,-3) = (x2 , y2 ),

C(6,2) = (x3, y3) are

three vertices of triangle ABC.

area( ∆ABC )

= 1/2|x1(y2-y3)+x2(y3 -y1)+x3(y1-y2)|

=1/2| -2[-3-2]+2[2-(-8)]+6[-8-(-3)]|

= 1/2|(-2)(-5)+2(10)+6(-5)|

= 1/2 | 10 + 20 - 30 |

= 1/2 | 30 - 30 |

= 1/2 × 0

= 0

Therefore ,

Area of the triangle ABC = 0

A( -2,-8), B(2,-3) and C(6,2) are lying

on a line .

A , B and C are collinear points .

I hope this helps you.

: )