Question

Show that A (4,1), B (5,-2) and c (6,-5) are collinear. [Note: slope of AB = slope of BC= slope of AC​

Answer 1

◇Given to prove :-

A=(4 ,1) B=(5, -2) C=(6 , -5) are collinear

◇Proof :-

☛If slope of AB = slope of BC = slope of AC are equal then the points are said to be collinear .So, Let's show their slopes are equal .

$\pink \bigstar\: \: \: \: \: \boxed{ \underline{ \sf \: Slope \: of \: AB \: = \frac{y_2 - y_1}{x_2 - x_1} }}$

A= (4, 1) = (x₁ , y₁)

B = (5, -2) = (x₂ , y₂)

Substituting the values,  

$\longrightarrow \sf \: slope \: of \:  AB=\dfrac{ - 2 - 1}{5 - 4}$

$\longrightarrow \sf \: slope \: of \:AB =\dfrac{ - 3}{  1}$

$\:  \:  \star \boxed{ \sf \: slope \: of \:A = -  3}$

 B =  (5, -2) = (x₁ , y₁)

C = (6, -5) =  (x₂ , y₂)

$\longrightarrow \sf \: slope \: of \:  BC=  \dfrac{ - 5 - ( - 2) }{6 - 5}$

$\longrightarrow \sf \: slope \: of \:  BC=  \dfrac{ - 5  + 2 }{1}$

$\longrightarrow \sf \: slope \: of \:  BC=  \dfrac{ - 3 }{1}$

$\star \boxed{ \sf \: slope \: of \:BC = - 3}$

A =  (4, 1) = (x₁ , y₁)

C = (6, -5) =  (x₂ , y₂)

$\longrightarrow \sf \: slope \: of \:  A=  \dfrac{ - 5 - 1}{6 - 4}$

$\longrightarrow \sf \: slope \: of \:  AC=  \dfrac{ - 6}{2}$

$\:  \:  \:  \:  \star \boxed{ \sf \: slope \: of \:AC =   - 3}$

Since,

✐ slope of AB = slope of BC = slope of AC

We can say that given points are collinear.

★Proved !

Answer 2

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

Given coordinates are

Coordinates of A = (4, 1)

Coordinates of B = (5, - 2)

Coordinates of C = (6, - 5)

We know,

Slope of a line joining two points (a, b) and (c, d) is given by

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

So, Let's find the slope of the line joining the points A (4, 1) and B(5, - 2).

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

$\rm \: Slope\:of\:AB \: = \: \dfrac{ - 3}{1}$

$\rm\implies \:Slope\:of\:AB \: = \: - \: 3 - - - (1)$

Now, Let's find the slope of line joining the points B(5, - 2) and C (6, - 5).

$\rm \: Slope\:of\:BC \: = \: \dfrac{ - 5 + 2}{6 - 5}$

$\rm \: Slope\:of\:BC \: = \: \dfrac{ - 3}{1}$

$\rm\implies \:Slope\:of\:BC \: = \: - \: 3 \: - - - (2) \:$

So, from equation (1) and (2), we concluded that

$\rm \: Slope\:of\:AB \: = \: Slope\:of\:BC$

We know,

Two lines having slope m and M are parallel iff M = m

So, using this, we get

$\rm\implies \:AB \: \parallel \: BC$

$\rm\implies \:A, \: B, \: C \: are \: collinear$

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Additional Information :-

Slope of line is defined as the tangent of an angle p which a line makes with positive direction of x axis measured in anti-clockwise direction and is denoted by m and is given by m = tan p

Two lines having slope m and M are perpendicular iff Mm = - 1.

If a line is parallel to x - axis or it self x - axis, then slope is 0.

If a line is parallel to y - axis or it self y - axis, then slope is not defined.

If line makes an acute angle with positive direction of x axis, then slope is positive.

If line makes an obtuse angle with positive direction of x axis, then slope is negative.