Question

Find the value of k if the point A (2, 3), B (4, k) and C (6, - 3) are collinear.

Answer 1

Given that the points A(2,3),B(4,k) and C(6,−3) are collinear.

As we know that if three points are collinear then they will lie of a same plane and thus will not form a triangle.

Therefore,

Area of triangle formed by these points will be 0

Therefore,

Now,

Area of △ formed by these points =

2

1

×

∣

∣

∣

∣

∣

∣

∣

∣

2

4

6

3

k

−3

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

Therefore,

∣

∣

∣

∣

∣

∣

∣

∣

2

4

6

3

k

−3

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=0

[2(k−(−3))−3(4−6)+1((−12)−6k)]=0

2k+6+6−12−6k=0

−4k=0

⇒k=0

Thus the value of k is 0.

Hence the correct answer is 0.

Answer 2

Step-by-step explanation:

If the points are collinear area will be 0.

A(2, 3), B (4, k) and C (6, - 3)

(X1,y1) (x2,y2). (x3,y3)

$area = \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3 (y1 - y2))$

0=½[2(k-(-3)+4(-3-3)+6(3-k)]

0(2)=[2(k+3)+4(-3-3)+6(3-k)]

0=[2k+6-24+18-6k]

0=-4k

0=k.