Question

Verify whether the following points are collinear
(1) (1,-1), (4,1),(-2,-3)
(i) (1,-1), (2,3), (2,0)
(ii) (1,-6), (3,-4), (4, -3)​

Answer 1

Answer:

0,5,5

0,5,2

-5,-1,1 Is the answer of this question

Answer 2

Answer:

1) ( 1, - 1 ), ( 4, 1 ), ( - 2, - 3 ) are collinear.

2) ( 1, - 1 ), ( 2, 3 ), ( 2, 0 ) are not collinear.

3) ( 1, - 6 ), ( 3, - 4 ), ( 4, - 3 ) are collinear.

Step-by-step-explanation:

We have given some points with their coordinates.

We have to verify whether the points are collinear or not.

We can use slope formula to verify the given points whether they are collinear or not.

$\pink{\sf\:Slope\:(\:m\:)\:=\:\dfrac{y_2\:-\:y_1}{x_2\:-\:x_1}}\sf\:\:\:-\:-\:-\:[\:Formula\:]$

Now,

1) ( 1, - 1 ), ( 4, 1 ), ( - 2, - 3 )

Let the points be A, B and C respectively with the above coordinates.

$\therefore\sf\:\bullet\:A\:\equiv\:(\:1\:,\:-1\:)\\\\\\\sf\:\bullet\:B\:\equiv\:(\:4\:,\:1\:)\\\\\\\sf\bullet\:C\:\equiv\:(\:-\:2\:,\:-\:3\:)\\\\\\\sf\:Now\:,\\\\\\\sf\:Slope\:of\:line\:AB\:=\:\dfrac{1\:-\:(\:-\:1\:)}{4\:-\:1}\\\\\\\implies\sf\xSlope\:of\:line\:AB\:=\:\dfrac{1\:+\:1}{3}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:AB\:=\:\dfrac{2}{3}}}\\\\\\\sf\:Slope\:of\:line\:BC\:=\:\dfrac{-\:3\:-\:1\:}{-\:2\:-\:4}\\\\\\\implies\sf\:Slope\:of\:line\:BC\:=\:\dfrac{\cancel{-}\:4}{\cancel{-}\:6}\\\\\\\implies\sf\:Slope\:of\:line\:BC\:=\:\cancel{\dfrac{4}{6}}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:BC\:=\:\dfrac{2}{3}}}$

Slope of line AB = BC and B is the common point.

$\therefore\underline{\sf\:The\:given\:points\:are\:collinear}$

$\rule{200}{1}$

2) ( 1, - 1 ), ( 2, 3 ), ( 2, 0 )

Let the points be L, M & N with the above coordinates.

$\therefore\bullet\sf\:L\:\equiv\:(\:1\:,\:-\:1\:)\\\\\\\bullet\sf\:M\:\equiv\:(\:2\:,\:3\:)\\\\\\\bullet\sf\:N\:\equiv\:(\:2\:,\:0\:)$

Now,

$\sf\:Slope\:of\:line\:LM\:=\:\dfrac{3\:-\:(\:-\:1\:)}{2\:-\:1}\\\\\\\implies\sf\:Slope\:of\:line\:LM\:=\:\dfrac{3\:+\:1}{1}\\\\\\\implies\sf\:Slope\:of\:line\:LM\:=\:\dfrac{4}{1}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:LM\:=\:4}}\\\\\\\sf\:Slope\:of\:line\:MN\:=\:\dfrac{0\:-\:3}{2\:-\:2}\\\\\\\implies\sf\:Slope\:of\:line\:MN\:=\:\dfrac{-\:3}{0}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:MN\:cannot\:be\:determined\:}}\\\\\\\sf\:Slope\:of\:line\:LM\:\neq\:Slope\:of\:line\:MN\\\\\\\therefore\underline{\sf\:The\:given\:points\:are\:not\:collinear}$

$\rule{200}{1}$

3) ( 1, - 6 ), ( 3, - 4 ), ( 4, - 3 )

Let the points be X, Y & Z having the above coordinates respectively.

$\therefore\bullet\sf\:X\:\equiv\:(\:1\:,\:-\:6\:)\\\\\\\bullet\sf\:Y\:\equiv\:(\:3\:,\:-\:4\:)\\\\\\\bullet\sf\:Z\:\equiv\:(\:4\:,\:-\:3\:)$

Now,

$\sf\:Slope\:of\:line\:XY\:=\:\dfrac{-\:4\:-\:(\:-\:6\:)}{3\:-\:1}\\\\\\\implies\sf\:Slope\:of\:line\:XY\:=\:\dfrac{-\:4\:+\:6}{2}\\\\\\\implies\sf\:Slope\:of\:line\:XY\:=\:\cancel{\dfrac{2}{2}}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:XY\:=\:1}}\\\\\\\sf\:Slope\:of\:line\:YZ\:=\:\dfrac{-\:3\:-\:(\:-\:4\:)}{4\:-\:3}\\\\\\\implies\sf\:Slope\:of\:line\:YZ\:=\:\dfrac{-\:3\:+\:4}{1}\\\\\\\implies\sf\:Slope\:of\:line\:YZ\:=\:\dfrac{1}{1}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:YZ\:=\:1}}$

Slope of line XY = Slope of line YZ and Y is the common point.

$\therefore\underline{\sf\:The\:given\:points\:are\:collinear}$