Question

using distance formula, show that points(-3,2),(1,-2)and(9,-10)are collinear
​

Answer 1

Solution :-

Let the points be A ( -3 , 2 ), B ( 1 , -2 ) and C ( 9 , -10 ).

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

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

        $= \sf \sqrt{(1+3)^2+(-2-2)^2} \\\\= \sqrt{4^2+(-4)^2} \\\\= \sqrt{16+16} \\\\=\sqrt{2 \times 16} \\\\= 4\sqrt{2} \ units$

$\bullet \sf \ BC = \sqrt{(9-1)^2+(-10-(-2))^2}$

        $= \sf \sqrt{(9-1)^2+(-10+2)^2} \\\\= \sqrt{8^2+(-8)^2} \\\\= \sqrt{64+64} \\\\= \sqrt{64 \times 2 } \\\\= 8\sqrt{2} \ units$

$\bullet \sf \ AC = \sqrt{(9-(-3))^2+(-10-2)^2}$

        $= \sf \sqrt{(9+3)^2+(-10-2)^2} \\\\= \sqrt{12^2+(-12)^2} \\\\= \sqrt{144+144} \\\\= \sqrt{2 \times 144} \\\\= 12 \sqrt{2} \ units$

AB + BC = $\sf 4 \sqrt{2} + 8 \sqrt{2}$

              $= \sf 12 \sqrt{2}$

           

∴ AB + BC = AC

∴ The given points are collinear.

Answer 2

Answer:

Given :

• (-3,2),(1,-2)and(9,-10)

To Find :

• show that points(-3,2),(1,-2)and(9,-10)are collinear

Solution :

Let A( x1 , y1 ) , B( x2 , y2 ) are the

two points ,

Distance between A and B

= AB

= √ ( x2 - x1 )² + ( y2 - y1 )²

Concept :

Distance formula is the formula to measure distance between two points

• In the case of polygons, distance formula between 2 points are distance = √(x2 - x1)²+(y2-y1)²

$\large{ \underline{ \bf Formula \: to \: be \: used : }}$

$\bf{Distance \: formula : } \sf If \:( x _{1} , y_{1}) \: and \: (x_{2} , y_{2})$

Substitute all Values :

Now calculating AB , BC and AC by using distance formula :

First calculate AB

$\sf :\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AB = \sqrt{(-2 - 2)^{2} + (1 + 3)^{2} } \\\\ \\\sf :\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AB = \sqrt{(-4)^{2} + (4)^{2} } \\ \\ \\ \sf : \implies\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AB = \sqrt{16+16} \\\\ \\ \sf : \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:AB = 4\sqrt{2}$

Calculate BC

$\sf :\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:BC = \sqrt{(-10+2)^{2} + (9 - 1)^{2} } \\\\\\\sf :\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:BC = \sqrt{(-8)^{2} + (8)^{2} }\\\\\\ \sf:\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: BC = \sqrt{64 + 64} \\\\ \\ \sf:\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:BC = 8\sqrt{2}$

Calculate AC

$\sf :\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:AC = \sqrt{(-10-2)^{2} +(9+3)^{2} } \\\\ \\ \sf :\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:AC = \sqrt{(-12)^{2} +(12)^{2} } \\\\ \\ \sf:\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AC = \sqrt{144 + 144} \\ \\\\ \sf:\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AC = 12\sqrt{2}$

Now Adding the Value of AB and BC :

How To Determine If Points Are Collinear ?

• Slope formula method to find that points are collinear. Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C, three pairs of points can be formed, they are: AB, BC and AC. If Slope of AB = slope of BC = slope of AC, then A, B and C are collinear points.

$\sf:\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AB + BC = 4\sqrt{2}+8\sqrt{2} \\\\\\ \sf:\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AB + BC = 12\sqrt{2} \\\\\\ \sf:\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AB + BC = AC$

• Hence he points are A(-3,2) , B(1,-2) and C(9, -10) collinear.