Question

show that the points (1,5);(2,3);(-2,-11) are collinear.

Answer 1

Hi!

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Distance Formula :

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

Let three points be

a ( 1 , 5 ) ; b (2, 3) ; c ( -2 , -11)

Find AB

$\sqrt{(2-1)^2 + (3 -5)^2 }$

$\sqrt{(1)^2 + (-2)^2}$

$\sqrt{1 + 4 }$

AB = $\sqrt{5}$

Find AC

$\sqrt{(-2 - 1)^2 +( -11 - 5 )^2}$

= $\sqrt{(-3)^2 + (-16)^2 }$

= $\sqrt{9 + 256 }$

AC = $\sqrt{265}$

Find BC

$\sqrt{(-2 - 2)^2 + (-11 -3 )^2 }$

=$\sqrt{(-4)^2 + (-14)^2 }$

= $\sqrt{16 + 196 }$

BC = $\sqrt{212}$

Here , AB + BC ≠ AC , BC + AC ≠ AB and AB + AC ≠ BC

From the above Points a , b and c  are not Collinear

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Thanks !!

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let A(1 , 5) , \: B(2 , 3) , \: C(-2 , -11)$

$\bf\huge AB= \sqrt{(2 - 1)^2 + (3 - 5)^2}$

$\bf\huge = \sqrt{1^2 + (-2)^2}$

$\bf\huge = \sqrt{1 + 4} = \sqrt{5}$

$\bf\huge BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2}$

$\bf\huge BC = \sqrt{(-4)^2 + (-14)^2}$

$\bf\huge BC = \sqrt{16 + 196}$

$\bf\huge BC = \sqrt{212}$

$\bf\huge AB = \sqrt{4\times 53} = 2\sqrt{53}$

$\bf\huge AC = \sqrt{(-2 - 1)^2 + ( - 11 - 5)^2}$

$\bf\huge AC = \sqrt{(-3)^2 + (- 16)^2}$

$\bf\huge AC = \sqrt{9 + 256} = \sqrt{265}$

$\bf\huge Hence \:A , B\: and \:C\: are\: not\: collinear$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$