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Find the values of k if the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear

Solution:

Given, the points A(k + 1, 2k) B(3k, 2k + 3) and C(5k - 1, 5k) are collinear.

We have to find the values of k.

To check for the points to be collinear, the area of the triangle must be zero.

The area of a triangle with vertices A (x₁ , y₁) , B (x₂ , y₂) and C (x₃ , y₃) is

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

Here, (x₁ , y₁) = (k + 1, 2k), (x₂ , y₂) = (3k, 2k + 3) and (x₃ , y₃) = (5k - 1, 5k)

Area of the triangle = 1/2[(k + 1)(2k + 3 - 5k) + 3k(5k - 2k) + (5k - 1)(2k - (2k + 3))] = 0

1/2[(k + 1)(3 - 3k) + 3k(3k) + (5k - 1)(-3)] = 0

3k - 3k² + 3 - 3k + 9k² - 15k + 3 = 0

By grouping,

9k² - 3k² - 15k - 3k + 3k + 3 + 3 = 0

6k² - 15k + 6 = 0

Dividing by 3,

2k² - 5k + 2 = 0

2k² - 4k - k + 2 = 0

2k(k - 2) - 1(k - 2) = 0

(2k - 1)(k - 2) = 0

Now, 2k - 1 = 0

2k = 1

k = 1/2

Also, k - 2 = 0

k = 2

Therefore, the values of k are 1/2 and 2.

✦ Try This: Find the values of k if the points (k, 2 - 2k), (-k + 1, 2k) and (-4 - k, 6 - 2k) are collinear.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 19

Summary:

The values of k if the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear are 1/2 and 2

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