Question

solve this question
no irrelevant answers​

Answer 1

Step-by-step explanation:

For proving collinearity, we need to prove AB + BC = AC

Distance formula is,

$\boxed{\sf \bf D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

Finding AB:

$\sf \bf : \longmapsto AB = \sqrt{(a-b)^2 + (b+c - c - a)^2}$

$\sf \bf : \longmapsto AB = \sqrt{(a-b)^2 + (b-a)^2}$

$\sf \bf : \longmapsto AB = \sqrt{2(a-b)^2}$

$\sf \bf : \longmapsto AB = \pm \sqrt2(a-b)$

Finding BC:

$\sf \bf : \longmapsto BC = \sqrt{(b-c)^2 + (c+a-a-b)^2}$

$\sf \bf : \longmapsto BC = \sqrt{(b - c)^2 + (c - b)^2}$

$\sf \bf : \longmapsto BC = \sqrt{2(b-c)}$

$\sf \bf : \longmapsto BC = \pm \sqrt2(b-c)$

Finding AC:

$\sf \bf : \longmapsto AC = \sqrt{(a-c)^2+ (b+c-a-b)^2}$

$\sf \bf : \longmapsto AC = \sqrt{(a-c)^2 + (c-a)^2}$

$\sf \bf : \longmapsto AC = \sqrt{2(a-c)^2}$

$\sf \bf : \longmapsto AC = \pm \sqrt2(a-c)$

Now checking the condition of collinearity

$\sf \bf : \longmapsto AB + AC = \pm \sqrt2 [a-b+b-c]$

$\sf \bf : \longmapsto AB + AC = \pm \sqrt2(a-c)$

Clearly, AB + BC = AC

This proves that A,B,C are collinear