Question

find k if the points A(4,6) B(8,2k) C(12,-6) are collinear​

Answer 1

The given points are said to be collinear if

Area of a triangle=Δ=

2

1

∣

∣

∣

∣

∣

∣

∣

∣

x

1

x

2

x

3

y

1

y

2

y

3

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=0

⇒

2

1

∣

∣

∣

∣

∣

∣

∣

∣

4

8

12

6

2k

r

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=0

⇒4(2k−r)−6(8−12)+1(8r−24k)=0

⇒8k−4r+24+8r−24k=0

⇒24+4r−16k=0

⇒6+r−4k=0

⇒−4k=−6−r

∴k=

4

6+r

Hope it's help to you buddy and Mark this answer as brainlist

Answer 2

Answer:

Mark me as brainlist now please

Step-by-step explanation:

A(4,6)

B(8,2k)

C(12,-6)

Given that A , B and C are collinear

They will lie on a same line i.e. They will not form triangle

AREA OF /\ABC = 0

1/2[x1 ( y2-y3) + x2(y3-y1) + x3(y1-y2)]=0

Here

x1 = 4 , y1 = 6

x2 = 8 , y2 = 2k

x3 =12 , y3 =-6

_________________

=> 1/2[ 4(2k -(-6)+8(-6-6)+ 12(6-2k)] = 0

=> 1/2 [4(2k+6) + 8(-12) + 12(6-2k)] = 0

=> 1/2 [ 8k + 24 - 96 + 72 - 24k] = 0

=> 1/2 [ - 16k + 24 - 24 ] = 0

=> - 16k = 0

=> k = 0

So , value of 'k' is 0..

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