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In geometry, collinear points are points that lie on the same single line. A plane's position is determined by a point. On a plane, we can mark any number of points.
- The two points are said to be collinear if two or more than two points are lying close or far to each other.
- The phrase "co-" denotes togetherness, while the word "linear" denotes the line.
- These two terms combine to form the word "collinear."
- Assume that if you were to mark three points on a piece of paper, you would have to write a single capital letter next to each one, such as A, B, and C, since we have to use capital letters to symbolise the points.
- When these three points are collinear, we may also draw a variety of forms that cross across them, such as a line, ray, or line segment.
- We can draw a triangle, circle, etc, with the help of collinearity of points.
- Only one circle, made from three non-collinear points, may be drawn.
- If points do not lie on a straight line, then points are said to be non-collinear points.
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Key Terms: Collinear Points, Collinearity of Points, Slope Formula Method, Area of Triangle Method, Straight Line, Non- Collinear Points
Collinear Points Definition
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Sets of points that are all on the same straight line are known as collinear points. Though not on separate lines, these points can lie on several planes.
- There is only one straight line that can be drawn using the sets of three collinear points.
- Two points may always be used to draw a straight line.
- Thus, it may be concluded that two points are always collinear.
- We verify three points of collinearity because uncertainty arises when there are more than two points.
Real life Example of Collinear PointsExamples from real life include the small and long clock arrows from collinear points, the moment the clock strikes six, the assembly hall students standing in a straight line, the trucks parking on the side of the road in a straight line, the food items on a single straight skewer, the points formed by the adjoining two walls, and the cars parked in a row. |
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| Important Concepts Related to Collinear Points | ||
|---|---|---|
| Distance between Two Points | Horizontal and Vertical Lines | Lines and Angles |
| Vertex | Transversal | Properties of Parallel Lines |
| Angle Formula | Obtuse Angle | Linear Pair of Angles |
How to prove if points are Collinear?
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The conditions that are used to prove the collinearity of three points are as follows:
Slope Formula Method
Lines can be parallel, perpendicular, intersecting, etc., in general. The line slopes in each of these instances have some relationship with one another.
- Two lines will coincide if they have the same slope and travel through the same location.
- Stated otherwise, given three points A, B, and C in the XY-plane, they will lie on a line; that is, three points are collinear if and only if the slopes of AB and BC are equal.
- Using the slope formula, we can get the above condition for the collinearity of three points: A, B, and C.
- Let's imagine that the coordinates of three points are (x1, y1), (x2, y2) and (x3, y3) respectively.
- AB's slope equals BC's slope, indicating that A, B, and C are collinear.
- AB plus BC equals AC and AC + BC equals AB, or AB + AC = BC.
- This situation is illustrated in the image below.
Slope Formula Method
Area of Triangle Method
Collinearity of three points can be proved with the help of the Area of Triangle Method. Three points are said to be collinear if the value of the area of the triangle formed by these three coordinates is equivalent to zero.
Proof of Collinearity of three Points using Area of Triangle MethodLet's imagine that the coordinates of three points are (x1, y1), (x2, y2) and (x3, y3) respectively. AB's slope equals BC's slope, indicating that A, B, and C are collinear. (y2 – y1)/ (x2 – x1) = (y3 – y2)/ (x3 – x2) (x3 – x2)(y2 – y1) = (x2 – x1)(y3 – y2) x3(y2 – y1) – x2(y2 – y1) = x2(y3 – y2) – x1(y3 – y2) Changing the phrases around, x2(y3 – y2) + x1(y2 – y3) + x2(y2 – y1) – x3(y2 – y1) = 0 x1(y2 – y3) + x2(y3 – y2 + y2 – y1) + x3(y1 – y2) = 0 x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
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Things to Remember
- The term "collinear points" refers to three or more points that are on the same straight line.
- Non-linear points are those that are situated in nonlinear places where a straight line cannot be established.
- To determine whether or not points are collinear, there are three primary methods.
- The slopes created by any two of the three collinear points are equal to the slopes formed by the other two.
- Any three collinear points will always have an area of zero in the triangle that they create.
- The distance formula is used to calculate the difference between three points that lie on the same line.
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Sample Questions
Ques. Find the value of p for which the points (p, -1), (2, 1) and (4, 5) are collinear? (3 marks)
Ans. Let the given points be:
- A(p, -1) = (x1, y1)
- B(2, 1) = (x2, y2)
- C(4, 5) = (x3, y3)
- Given that A, B, and C are collinear.
- Slope of AB = Slope of BC
- (y2 – y1)/(x2 – x1) = (y3 – y2)/(x3 – x2)
- Substituting the values of coordinates of given points,
- (1 + 1)/(2 – p) = (5 – 1)/(4 – 2)
- 2/(2 – p) = 4/2
- 2/(2 – p) = 2
- 2 – p = 1
- p = 2 – 1
- p = 1
- Hence, the value of p is 1.
Ques. Using the equation method, check the collinearity of the points A(7, -2), B(2, 3) and C(-1, 6)? (3 marks)
Ans. We know that the equation of a line passing through the points (x1, y1) and (x2, y2) is:
- y – y1 = [(y2 – y1) /(x2 – x1)] (x – x1)
- Let A(7, -2) = (x1, y1) and B(2, 3) = (x2, y2).
- So, the equation of a line passing through the points A(7, -2) and B(2, 3) is given by:
- y + 2 = [(3 + 2)/(2 – 7)] (x – 7)
- y + 2 = (5/-5) (x – 7)
- y + 2 = -x + 7
- x + y + 2 – 7 = 0
- x + y – 5 = 0
- Now, substituting the point C(-1, 6) in the above equation,
- -1 + 6 – 5 = 0
- 0 = 0
- Thus, the third point satisfies the equation of the line passing through the two of given three points.
- Therefore, the given points A, B and C are collinear.
Ques. Show that the three points P(2, 4), Q(4, 6) and R(6, 8) are collinear? (3 marks)
Ans. If the three points P(2, 4), Q(4, 6) and R(6, 8) are collinear, then slopes of any two pairs of points, PQ, QR & PR will be equal.
- Now, using slope formula we can find the slopes of the respective pairs of points, such that;
- Slope of PQ = (6 – 4)/ (4 – 2) = 2/2 = 1
- Slope of QR = (8 – 6)/ (6 – 4) = 2/2 = 1
- Slope of PR = (8 – 4) /(6 – 2) = 4/4 = 1
- As we can see, the slopes of all the pairs of points are equal.
- Therefore, the three points P, Q and R are collinear.
Ques. Prove that the points (2, 4), (4,6), (6,8) are collinear using the Slope Formula? (3 marks)
Ans. Let P (2,4), Q(4,6), and R(6,8) be the given points.
- The slope of PQ =6-4 /4-2=1
- Slope of QR=8-6 / 6-4=1
- Slope of PR=8-4 / 6-2=1
- Slope of line PQ= Slope of line QR= Slope of line PR
- Hence, A, B, and C are Collinear.
Ques. Prove that points A(5, -2), B(4, -1) and C(1, 2) are collinear points using the Distance Method? (3 marks)
Ans. Distance between any two points (x1, y1) and (x2, y2) is
- d = √[(x2 - x1)2 + (y2 - y1)2]
- To find the lengths AB, BC and AC using the formula,
- AB = √[(4 - 5)2 + (-1 + 2)2]
- AB = √[(-1)2 + (1)2]
- AB = √[1 + 1]
- AB = √2
- BC = √[(1 - 4)2 + (2 + 1)2]
- BC = √[(-3)2 + (3)2]
- BC = √[9 + 9]
- BC = √18
- BC = 3√2
- AC = √[(1 - 5)2 + (2 + 2)2]
- AC = √[(-4)2 + (4)2]
- AC = √[16 + 16]
- AC = √32
- AC = 4√2
- Therefore, AB + BC = √2 + 3√2 = 4√2 = AC
- Thus, AB + BC = AC
- This proves that given three points A, B, and C are collinear.
Ques. Prove that the given three points (4, 4), (-2, 6), and (1, 5) are collinear points using the Slope Formula Method? (3 marks)
Ans. Formula: m = (y2 - y1)/(x2-x1)
Step 1 : AB’s Slope : (x1, y,) ==> (4 , 4) and (x2, y2) ==> (-2 , 6)
- m = (6 - 4) / (-2 - 4)
- 2/(-6)
- -1/3
Step 2 : BC’s Slope : (x1, y1) ==> (-2, 6) and (x2, y2) ==> (1, 5)
- m = (5 - 6) / (1 - (-2))
- (-1 )/(1 + 2)
- -1/3
Step 3 : Slope of ‘AB’ = Slope of ‘BC’
Hence, the given points are collinear
Ques. What is the equation of the circle passing through the points A(2, 0), B(-2, 0) and C(0, 2)? (3 marks)
Ans. Consider the general equation of circle:
- x2 + y2 + 2gx + 2fy + c = 0….(i)
- Substituting A(2, 0) in (i),
- (2)2 + (0)2 + 2g(2) + 2f(0) + c = 0
- 4 + 4g + c = 0….(ii)
- Substituting B(-2, 0) in (i),
- (-2)2 + (0)2 + 2g(-2) + 2f(0) + c = 0
- 4 – 4g + c = 0….(iii)
- Substituting C(0, 2) in (i),
- (0)2 + (2)2 + 2g(0) + 2f(2) + c = 0
- 4 + 4f + c = 0….(iv)
- Adding (ii) and (iii),
- 4 + 4g + c + 4 – 4g + c = 0
- 2c + 8 = 0
- 2c = -8
- c = -4
- Substituting c = -4 in (ii),
- 4 + 4g – 4 = 0
- 4g = 0
- g = 0
- Substituting c = -4 in (iv),
- 4 + 4f – 4 = 0
- 4f = 0
- f = 0
- Now, substituting the values of g, f and c in (i),
- x2 + y2 + 2(0)x + 2(0)y + (-4) = 0
- x2 + y2 – 4 = 0
- Or
- x2 + y2 = 4
- This is the equation of the circle passing through the given three points A, B and C.
Ques. Show that the three points P(2, 2), Q(4, 2) and R(6, 6) are collinear? (3 marks)
Ans. If the three points P(2, 2), Q(4, 2) and R(6, 6) are collinear, then slopes of any two pairs of points, PQ, QR & PR will be equal.
- Now, using slope formula we can find the slopes of the respective pairs of points, such that;
- Slope of PQ = (2 – 2)/ (4 – 2) = 0
- Slope of QR = (6 - 2)/ (6 – 4) = 4/2 = 2
- Slope of PR = (6 – 2) /(6 – 2) = 4/4 = 1
- As we can see, the slopes of all the pairs of points are not equal.
- Therefore, the three points P, Q and R are not collinear.
Ques. Show that points A(2, 1), B(4, 3), and C(6, 10) are collinear points using the Area of triangle Method? (3 marks)
Ans. Given points A, B and C of triangle with their coordinates (2,1) (4, 3) and (6,10)
- Area of triangle (⧍ABC) = 0
- 1/2[x1(y2 -y3 ) + x2 (y3-y1 ) + x3 (y1 -y2 )] = 0
- 1/2[2(3 -1 ) + 4 (10 - 3 ) + 6 (1 -10)]
- 1/2[4+28-54] = 0
- Area not equal to 0
- Hence, the points A, B and C are not collinear.
Ques. Show that the three points P(1, 5), Q(4, 3) and R(6, 5) are collinear? (3 marks)
Ans. If the three points P(1, 5), Q(4, 3) and R(6, 5) are collinear, then slopes of any two pairs of points, PQ, QR & PR will be equal.
- Now, using slope formula we can find the slopes of the respective pairs of points, such that;
- Slope of PQ = (3 – 5)/ (4 – 1) = -2/3
- Slope of QR = (5 - 3)/ (6 – 4) = 2/2 = 1
- Slope of PR = (5 – 5) /(6 – 1) = 0
- As we can see, the slopes of all the pairs of points are not equal.
- Therefore, the three points P, Q and R are not collinear.
Ques. Show that points A(1, 1), B(1, 3), and C(5, 10) are collinear points using the Area of triangle Method? (3 marks)
Ans. Given points A, B and C of triangle with their coordinates (2,1) (4, 3) and (6,10)
- Area of triangle (⧍ABC) = 0
- 1/2[x1(y2 -y3 ) + x2 (y3-y1 ) + x3 (y1 -y2 )] = 0
- 1/2[1(3 -1 ) + 1 (10 - 3 ) + 5 (1 -10)]
- 1/2[2+7-45] = 0
- Area not equal to 0
- Hence, the points A, B and C are not collinear.
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