(1,-1) (4,1) (-2,-3) are collinear or not verify
Answers
Step-by-step explanation:
Let,
A(1,-1), B(4,1), C(-2,-3)
Now Consider,
A(1,-1), B(4,1)
x1 = 1, x2 = 4; y1 = -1, y2 = 1,
W.K.T,
___________________________________
The distance between 2 points = √(x2-x1)^2 + (y2-y1)^2
___________________________________
= √[(4-1)^2 + (1-(-1)^2]
= √(4-1)^2 + (1+1)^2
= √(3)^2 + (2)^2
= √(9) + (4)
= √13
:. AB = √13 units.
Now Consider,
B(4,1), C(-2,-3)
x1 = 4, x2 = -2, ; y1 = 1, y2 = -3
W.K.T,
___________________________________
The distance between 2 points = √(x2-x1)^2 + (y2-y1)^2
___________________________________
= √(-2-4)^2 + (-3-1)^2
= √(-6)^2 + (-4)^2
= √(36) + (16)
= √52
= 2√26
:. BC = 2√26 units.
Now Consider,
A(1,-1),C(-2,-3)
x1 = 1, x2 = -2, y1 = -1, y2 = -3
W.K.T,
___________________________________
The distance between 2 points = √(x2-x1)^2 + (y2-y1)^2
___________________________________
= √[(-2-1)^2 + (-3-(-1)^2]
= √(-2-1)^2 + (-3+1)^2
= √(-3)^2 + (-2)^2
= √(9) + (4)
= √13
:.AC = √13 units.
Now,
BC = AB + AC
2√26 = √13 + √13
2√26 = 2√26
AB + AC = BC So, it is collinear.
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