Question

A(-4,4), K(-2, 5/2), N(4,-2)Determine whether the given points are collinear or not.

Answer 1

Answer - Yes. 

Explanation -

Let the Points A(-4,4), K(-2, 5/2), N(4,-2) be A(x₁, y₁), K(x₂,y₂), N(x₃,y₃).

Let us first find the Slope of AK,

∵ m = $\frac{y_2 - y_1}{x_2 - x_1}$
∴ m = (5/2 - 4)/(-2 + 4)    
        = (-3/2)/2  
        = -3/4

Now For th Slope of KN, 

m = $\frac{y_3 - y_2}{x_3 - x_2}$   
    = (-2 - 5/2)/(4 + 2)  
    = (-9/2)/6
    = -9/12
    = -3/4


Since, the Slope of both the lines AK and KN are same therefore, Points are Collinear.


Hope it helps.

Answer 2

if three points are collinear:
1) Then slope of line is same,when we check taking two points at the time .
2) The area of triangle formed by these three lines is zero.(because we know that collinear points can not formed triangle)
Method one is discuss in another answer.
Method 2:
Area of triangle formed by three points (x1,y1),(x2,y2),(x3,y3),here A,K and N are these points respectively
$\frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| \\ = \frac{1}{2} | - 4( \frac{5}{2} + 2) + ( - 2)( - 2 - 4) + 4(4 - \frac{5}{2}) | \\ = \frac{1}{2} | - 4( \frac{9}{2}) - 2( - 6) + 4( \frac{8 - 5}{2}) | \\ = \frac{1}{2} | -18 + 12 + 6| \\ = \frac{1}{2} |18 - 18| \\ = \frac{1}{2} \times 0 \\ = 0 \\ so \: these \: given \: points \: are \: collinear$
hope you understand well.