Question

Check whether (5,-2) (6, 4)and (7,-2) are Collinear​

Answer 1

$\large\underline{\sf{Solution-}}$

Given points are ( 5, - 2 ), ( 6, 4 ) and ( 7, -2 )

Let assume that,

Coordinates of A be ( 5, - 2 )

Coordinates of B be ( 6, 4 )

Coordinates of C be ( 7, - 2 )

Let we use slope method to check whether given 3 points are collinear or not.

We know, 3 points A, B and C are collinear iff Slope of AB = Slope of BC.

Also, we know that, slope of line segment joining the points A ( a, b ) and B ( c, d ) is

$\red{ \boxed{ \sf{ \:Slope \: of \: AB \: = \: \frac{d - b}{c - a} \: }}}$

Thus,

Slope of line segment AB joining the points A ( 5, - 2 ), B ( 6, 4 ) is

$\rm :\longmapsto\:Slope \: of \: AB \: = \: \dfrac{4 - ( - 2)}{6 - 5} = \dfrac{6}{1} = 6$

And,

Slope of line segment BC joining the points B ( 6, 4 ) and ( 7, - 2 ) is

$\rm :\longmapsto\:Slope \: of \: BC \: = \: \dfrac{ - 2 - 4}{7- 6} = \dfrac{ - 6}{1} = - \: 6$

So, we concluded that

$\rm :\longmapsto\:Slope \: of \: AB \: \ne \: Slope \: of \: BC$

$\bf\implies \:A, \: B, \: C \: are \: not \: collinear$

Alternative Method :-

Using Area of triangle

We know, three points A, B and C are collinear iff area of triangle ABC = 0

Now, Area of triangle is given by

$\sf \ Area =\dfrac{1}{2}\bigg| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|$

So, here on substituting the values, we get

$\sf \ Area =\dfrac{1}{2}\bigg| 5(4 + 2) - 2( - 2 + 2) + 7( - 2-4)\bigg|$

$\sf \ Area =\dfrac{1}{2}\bigg| 5(6) - 2(0) + 7( -6)\bigg|$

$\sf \ Area =\dfrac{1}{2}\bigg|30 - 42\bigg|$

$\sf \ Area =\dfrac{1}{2}\bigg| - 12\bigg|$

$\sf \ Area =\dfrac{1}{2}\bigg|12\bigg|$

$\sf \ Area =6 \: \ne \: 0$

• Hence, Points are not collinear.