Solved Examples on Collinear Points

Example 1: Show that points A(5, -2), B(4, -1) and C(1, 2) are collinear points using the Distance Method.

Solution:

The points A, B and C are collinear,

if, (Distance from A to B) + (Distance from B to C) = (Distance from C to A)

By using Distance method, we can determine the distance between two points.

Here x₁= 5, y₁= -2, x₂ = 4, y₂= -1, x₃ = 1, y₃= 2

Distance between AB = √{(x₂ -x₁ )² + (y₂ -y₁ )² }

⇒ Distance between AB = √{(4 -5)² + (-1 -(-2))²}

⇒ Distance between AB = √2

Distance between BC = √{(x₃ -x₂ )2 + (y₃ – y₂ )²}

⇒ Distance between BC = √{(1 -4 )² + (2 -(-1))²}

⇒ Distance between BC = √18

⇒ Distance between BC = 3√2

Distance between AC = √{(x₃ -x₁ )²+(y₃ -y₁)²}

⇒ Distance between AC = √{(1 -5 )²+(2 -(-2))²}

⇒ Distance between AC = √32

⇒ Distance between AC = 4√2

Therefore, AB + BC = √2 + 3√2 = 4√2

Thus, AB + BC = AC

Hence, all given three points A, B, and C are collinear.

Example 2: Find A(2, 3), B(4, 7) and C(6, 11) are collinear points using slope method.

Solution:

Given the coordinates of point A, B and C are (2 ,3 ), (4 ,7) and (6 ,11 )

The points A, B and C are collinear,

if, slope of line AB = slope of line BC= slope of line CA

⇒ m_AB =m_BC = m_CA

(y₂ -y₁ )/(x₂ -x₁ ) = (y₃ -y₂ )/(x₃ -x₂ )=(y₃ -y₁ )/(x₃ -x₁ )

Here, x₁= 2 y₁= 3, x₂= 4, y₂= 7, x₃= 6, y₃= 11

⇒ (7 -3 )/(4 -2 ) = (11 -7 )/(6 -4)=(11 -3 )/(6 -2)

⇒ 4/2 = 4/2 =8/4

⇒ 2 = 2 = 2

Hence, slopes of all three points are equal.

Therefore, the three points A, B and C are collinear.

Example 3: Show that points A(2, 3), B(4, 7), and C(6, 11) are collinear points using the Area of triangle Method.

Solution:

Given points A, B and C of triangle with their coordinates (2,3) (4,7 ) and (6,11 )

Area of triangle (⧍ABC) = 0

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

⇒ 1/2[2(7 -11 ) + 4 (11-3 ) + 6 (3 -7)]

⇒ 1/2[-8+32-24] = 0

⇒ Area = 0

Hence, the points A, B and C are collinear.

Example 4: If the collinear points are given (a,0), (0,b) and (1,1). Then find out the value of (1/a + 1/b)².

Solution:

Given points A, B and C of triangle with their coordinates (a,0) (0,b) and (1,1)

Area of triangle (⧍ABC) = 0

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

⇒ x₁(y₂ – y₃ ) + x₂(y₃ – y₁ ) + x₃(y₁ – y₂ ) = 0

⇒ a(b-1) + 0(1-0) + 1(0-b) =0

⇒ ab – a – b=0

⇒ ab = a +b

⇒ ab/a+b =1

⇒ 1/a + 1/b = 1

⇒ (1/a + 1/b)² = 1

Collinear Points are sets of three or more than three points that lie in a straight line. In simple words, if three or more points are collinear, they can be connected with a straight line without any change in slope.

In this article, we will discuss the concept of collinear points, collinear point definition, collinear point meaning, and properties. We will also know how to determine the three points collinearity by different methods. Further, we will also solve various examples and provide practice questions for a better understanding of the concept of this article.