Question

Using Distance formula prove that A (16, -8), B (4, -3) and C (-8,2) are collinear.

Answer 1

Solution :-

Given :-

Points are A ( 16 , -8 ), B ( 4 , -3 ) and C ( -8 , 2 ).

First let's find AB, BC and AC.

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

$\bullet \sf \ AB = \sqrt{(4-16)^2+(-3-(-8))^2}$

        $= \sf \sqrt{(4-16)^2+(-3+8)^2} \\\\= \sqrt{(-12)^2+5^2} \\\\= \sqrt{144+25} \\\\= \sqrt{169 } \\\\= 13 \ units$

$\bullet\sf \ BC = \sqrt{(4-(-8))^2+(-3-2)^2}$

        $= \sf \sqrt{(4+8)^2+(-3-2)^2} \\\\= \sqrt{12^2+(-5)^2} \\\\= \sqrt{144+25} \\\\= \sqrt{169} \\\\= 13 \ units$

$\bullet \sf \ AC = \sqrt{(-8-16)^2+(2-(-8))^2}$

        $= \sf \sqrt{(-8-16)^2+(2+8)^2} \\\\= \sqrt{(-24)^2+10^2} \\\\= \sqrt{576+100} \\\\= \sqrt{676} \\\\= 26 \ units$

AB + BC = 13 + 13

              = 26

              = AC

AB + BC = AC

∴ The given points are collinear.