Question

05. Find value of k if (-3, 11), (6, 2) ad (k. 4)
ane colliner points.​

Answer 1

Answer:

K=4

Step-by-step explanation:

Collinear points means that that the area will be zero.

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

       ⇒ 1/2×[-3(2-4) + 6(4-(-3)) + k(-3-6)] = 0

       ⇒ 1/2×[-6 + 42 - 9k] = 0

       ⇒ 1/2×[36-9k] = 0

       ⇒ 18 - 4.5k = 0

       ⇒ 4.5k = 18

       ⇒ k = 4

Answer 2

Answer:

The value of k is 4.

Step-by-step explanation:

ATQ, the points (-3, 11), (6, 2) & (k, 4) are collinear points.

We know that when three collinear points are connected, the area enclosed will be 0 square units.

Therefore, we'll use the area of a triangle (formed by coordinates) formula to find out the value of k.

$\boxed{\boxed{\sf Area \ of \ \triangle = \frac{1}{2} \ \bigg( x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right) \bigg)}}$

x₁  $\longmapsto$ -3

x₂  $\longmapsto$ 6

x₃  $\longmapsto$ k

y₁  $\longmapsto$ 11

y₂  $\longmapsto$ 2

y₃  $\longmapsto$ 4

⇒ Area of Δ = 0

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

⇒ 0 = ¹/₂ (-3(2 - 4) + 6(4 - 11) + k(11 - 2))

⇒ 0 = ¹/₂ (-3(-2) + 6(-7) + k(9))

⇒ 0 = ¹/₂ (6 - 42 + 9k)

⇒ 0 × 2= - 36 + 9k

⇒ 0 = 9k - 36

⇒ 9k = 36

⇒ k =  36/9

⇒ k = 4

∴ The value of k is 4.