Question

using distance formula show that the points (1 5) (2 4) and (3 3) are collinear​

Answer 1

Solution :

Let the points be A(1 , 5), B(2 , 4) and C(3 , 3).

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

$\bullet\sf \ AB = \sqrt{(2-1)^2+(4-5)^2}$

$\:\:\:\qquad= \sf \sqrt{1^2+(-1)^2}$

$\:\:\:\qquad= \sf \sqrt{1+1}$

$\:\:\:\qquad= \sf \sqrt{2} \ units$

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

$\:\:\:\qquad= \sf \sqrt{1^2+(-1)^2}$

$\:\:\:\qquad= \sf \sqrt{1+1}$

$\:\:\:\qquad= \sf \sqrt{2} \ units$

$\bullet\sf \ AC = \sqrt{(3-1)^2+(3-5)^2}$

$\:\:\:\qquad= \sf\sqrt{2^2+(-2)^2}$

$\:\:\:\qquad= \sf \sqrt{4+4}$

$\:\:\:\qquad= \sf \sqrt{8}$

$\:\:\:\qquad= \sf 2 \sqrt{2} \ units$

AB + BC = √2 + √2

= 2√2 units

= AC

∴ AB + BC = AC

Hence, the points A( 1 , 5 ), B(2 , 4 ) and C( 3 , 3 ) are collinear.

Answer 2

Solution :-

• Let the points be A(1 , 5) , B(2 , 4) , C(3 , 3) .

$\boxed{\bf{Distance~formula~=~ \sqrt{( x_{2} } - x_{1})² + ( y_{2} - x_{1})²}}$

• ${\sf{AB}}$ ${\sf{ = \sqrt{(2 - 1)² + (4 - 5)²} }}$

$\: \: \:\:\:\:\:\: \: \:$ ${\sf{= \sqrt{1² + ( - 1)²}}}$

$\: \: \:\:\:\:\:\: \:\:$ ${\sf{ = \sqrt{1 + 1} }}$

$\: \:\: \:\:\:\:\:\: \:$ ${\sf{= \sqrt{2}\:units}}$

• ${\sf{BC}}$ ${\sf{= \sqrt{(3 - 2)² + (3 - 4)²}}}$

$\: \: \:\:\:\:\:\:\:\:$ ${\sf{= \sqrt{1² + ( - 1)²} }}$

$\: \:\: \: \:\:\:\:\:\:$ ${\sf{ = \sqrt{1 + 1} }}$

$\: \:\:\: \:\:\:\:\: \:$ ${\sf{= \sqrt{2}\:units}}$

• ${\sf{AC}}$ ${\sf{ = \sqrt{(3 - 1)² + ( - 5)²}}}$

$\: \: \:\:\:\:\:\:\: \:$ ${\sf{ = \sqrt{2² + ( - 2)²} }}$

$\: \:\: \:\:\:\:\:\: \:$ ${\sf{ = \sqrt{4 + 4}}}$

$\: \: \:\:\:\:\:\: \:\:$ ${\sf{ = \sqrt{8} }}$

$\: \:\:\:\:\: \:\:\:\:$ ${\sf{=\sqrt{2} \:units}}$

AB + BC ${\sf{ =\sqrt{2} + \sqrt{2}}}$

$\: \:\:\: \:\:\:\:\:\: \:\:\:\:\:\:\: \:$ ${\sf{= \:2 \sqrt{2} \: units}}$

$\: \:\:\: \:\:\:\:\:\:\:\:\:\:\:\:\: \:$ ${\sf{=\:AC}}$

.°. AB + BC = AC

Hence,

• The points A(1 , 5) , B(2 ,4) and C(3 , 3) are collinear. $\large{\bf\green{✓}}$