Question

find the value of k if the co ordinates of the points A(2,-2) B(-4,2) and C(-7,k) are collinear​

Answer 1

Answer:

x1=2,x2=-4,x3=-7

y1=-2,y2=2,y3=k

all 4G 44

Step-by-step explanation :

ar of triangle =

1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))

0=1/2(2(2-k)+(-4)(k+2)+(-7)(-2-2))

0-1/2(4-2k-4k-8+28)

0-1/2(-6k+24)

0=-3k+12

3k=12

k=12/3

k=4

Answer 2

Answer:

The value of k = 1  

Step-by-step explanation:

Given data  

A ( 2,-2) , B (-4, 2) and C (-7, K)  are Collinear points

collinear points are the points which are lies on same line

⇒ KA, B and C are lies on a same line then AB + BC = AC

AB = distance between A (2, -2) and B (-4, 2)

the distance between 2 points  ( x, y ) and ( x₁, y₁)  = $\sqrt{(x_{1} - x)^{2} + (y_{2} -y)^{2} }$  

⇒ AB =  $\sqrt{(-4 -2)^{2} + (2 -(-2))^{2} }$

         =  $\sqrt{(-6)^{2} + (4)^{2} }$ = $\sqrt{36+16 }$  = $\sqrt{64} = 8$  

⇒ BC = distance between B (-4, 2) , C (-7, k)

         =  $\sqrt{ (-7- (-4))^{2} +( k-2)^{2} }$  

         =  $\sqrt{(-7+4)^{2}+( k-2)^{2} }$  =  $\sqrt{(-3)^{2} + (k-2)^{2} }$  = $\sqrt{9 +(k-2)^{2} }$

⇒ AC = distance between A(2,-2) and C(-7, K)

          = $\sqrt{(-7-2)+ (K-(-2))^{2} }$  = $\sqrt{(-9)^{2} + (K+2)^{2} }$  

         = $\sqrt{81 +( K+2)^{2} }$  

⇒ AB + BC = AC

⇒  8 + $\sqrt{9+ (K-2)^{2} }$  =  $\sqrt{81 +(K+2)^{2} }$    

⇒ 64 + 9 + $(k-2)^{2}$   = 81 + $(k+2)^{2}$    

⇒ 64 + 9 + $k^{2}$ + 4 - 4k = 81 + $k^{2}$ + 4 + 4k

⇒ -8k = 81 - 64 -9

⇒ - 8k = 8

⇒  k =1