Collinearity of Three Points: Condition & Equation

A group of points is collinear in Euclidean geometry if they all lie in the same line, whether they are far away, close together, or form a ray, a line, or a line segment. The term ‘collinear’ comes from a Latin word that means ‘together and ‘line.’ As a result, collinear refers to three or more points on a plane that are all on the same straight line.

The collinearity of a set of points is the property of the points lying on a single line in geometry. A set of points with this feature is said to be collinear. Points aligned in a line or a row are generally referred to as lines or rows.

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Condition of Collinearity of Three Points

If the triangle area produced by the three points is zero, the three points are collinear. In the area of the triangle formula, substitute the coordinates of the three points that are provided. The provided points are collinear if the answer for the triangle’s area is zero.
If the slope of any two pairs of points is the same, three points are collinear.

Check for Collinearity of Three Points

There are two approaches for determining whether the three points are collinear: the slope formula method and the area of triangle formula method.

The slope formula method: The slope of a straight line passing through two given fixed points:

Let \(P\left(x_{1}, y_{1}\right)\) and \(Q\left(x_{2}, y_{2}\right)\) be any two fixed points. It is required to find the slope of the line through \(P\) and \(Q\).

As is clear from the diagram below, the slope of the line passing through \(P\) and \(Q\) is

Slope \(=m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(=\frac{\text { difference of ordinates of the given points }}{\text { difference of their abscissae }}\)

If the slope of any two pairs of points is the same, the three points are collinear.

With three points \(A, B\), and \(C\), three pairs of points can be formed; they are \(A B, B C\) and \(A C\)

If the slope of \(A B=\) slope of \(B C=\) slope of \(A C\), then \(A, B\), and \(C\) are collinear points.

Example: Prove that the three points \(A(2,4), B(4,6)\), and \(C(6,8)\) are collinear.
Solution: If the three points \(A(2,4), B(4,6)\), and \(C(6,8)\) are collinear, then the slopes of any two pairs of points will be equal.
That is,
Slope of \(A B=\frac{6-4}{4-2}=1\)
Slope of \(B C=\frac{8-6}{6-4}=1\)
Slope of \(C A=\frac{4-8}{2-6}=1\)
Thus, slope of \(A B=\) Slope of \(B C=\) Slope of \(C A\).
Hence, proved.

Area of triangle formula method: If three distinct points \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) are collinear, then we cannot form a triangle because there will be no altitude (height) for such a triangle. Therefore, three points \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) will be collinear if the area of \(\triangle A B C=0\).

That is, \(\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|\) or

\(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}=x_{1} y_{3}+x_{2} y_{1}+x_{3} y_{2}\)

Similarly, if the area of \(\triangle A B C\) is zero, the three points lie on the same straight line. Thus, three distinct points \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) will be collinear if and only if the area of \(\triangle A B C=0\).

Example: Find the value of \(k\) if the points \(P(2,3), Q(4, k)\), and \(R(6,-3)\) are collinear.
Solution: The triangle area generated by the points mentioned above must be zero because they are collinear.
\(\Rightarrow \frac{1}{2}|2(k+3)+4(-3-3)+6(3-k)|=0\)
\(\Rightarrow \frac{1}{2}(-4 k)=0\)
\(\Rightarrow k=0\)
Therefore, \(k=0\).
Let us verify the answer.
Area of \(\Delta P Q R=\frac{1}{2}|2(0+3)+4(-3-3)+6(3-0)|=0\)

Number of Circles That Can be Drawn Through Three Collinear Points

There is no way to draw a circle through three collinear points. A straight line, on which the three points lie, is the only figure drawn through them.

Circle through three collinear points: Consider the following three collinear points \(A, B\), and \(C\). Is it possible to draw a circle that connects these three points? Consider it. We can not if the points are collinear?

Collinear points are those that are on the same line or in the same direction. If we create a circle using these collinear points, the result will be that the circle only touches two points, while the third point may observe either inside or outside of the circle. The circle in this scenario never touches all three points.

Equation of Plane Passing Through Three Non-collinear Points

Non-collinear points are those that do not all lie on the same line, as the name implies. We know that an infinite number of planes can pass through a given vector that is perpendicular to it, but that there will always be one and only one plane that is perpendicular to the vector and passes through a particular point based on our knowledge from previous lessons.

The equation of a plane travelling through three non-collinear points will be the topic of this session. Both the Vector and Cartesian forms of equations will be examined.

Vector form: Consider three non-collinear points \(P, Q\), and \(R\) lying on a plane, whose position vectors are given by \(\vec{a}, \vec{b}\), and \(\vec{c}\), as illustrated in the diagram below.

The \(\overrightarrow{P Q}\) and \(\overrightarrow{P R}\) vectors are in the same plane. \(\overrightarrow{P Q} \times \overrightarrow{P R}\) denotes the perpendicular vector to the plane containing the points \(P, Q\), and \(R\). If it is the position vector of any point \(A\) in the plane containing \(P, Q\), and \(R\), then the equation of the plane passing through \(P\) and perpendicular to the vector \(\overrightarrow{P Q} \times \overrightarrow{P R}\) is given by using the vector equation of a plane as mentioned above, the equation of the plane passing through \(P\) and perpendicular to the vector \(\overrightarrow{P Q} \times \overrightarrow{P R}\) is given by \((\vec{r}-\vec{a}) \cdot(\overrightarrow{P Q} \times \overrightarrow{P R})=0\)

Also, as shown in the diagram above, and. By substituting these numbers in the equation above, we get

\((\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]=0\)

This is the equation of a plane in vector form that passes through three non-collinear points.

Cartesian form: Assume that the coordinates of the points \(P, Q\), and \(R\) are \(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)\), and \(\left(x_{3}, y_{3}, z_{3}\right)\) respectively, to translate the above equation into a Cartesian system. Point \(As\) coordinates should also be \(x, y\), and \(z\).

\(\overrightarrow{P A}=\left(x-x_{1}\right) \hat{i}+\left(y-y_{1}\right) \hat{j}+\left(z-z_{1}\right) \hat{k}\)

\(\overrightarrow{P Q}=\left(x_{2}-x_{1}\right) \hat{i}+\left(y_{2}-y_{1}\right) \hat{j}+\left(z_{2}-z_{1}\right) \hat{k}\)

\(\overrightarrow{P R}=\left(x_{3}-x_{1}\right) \hat{i}+\left(y_{3}-y_{1}\right) \hat{j}+\left(z_{3}-z_{1}\right) \hat{k}\)

We can now utilise the Vector form to replace the vectors listed above. This would allow us to present the equation in determinant form, as seen below.

\(\left| {\begin{array}{*{20}{c}} {\left( {x – {x_1}} \right)}&{\left( {y – {y_1}} \right)}&{\left( {z – {z_1}} \right)}\\ {\left( {{x_2} – {x_1}} \right)}&{\left( {{y_2} – {y_1}} \right)}&{\left( {{z_2} – {z_1}} \right)}\\ {\left( {{x_3} – {x_1}} \right)}&{\left( {{y_3} – {y_1}} \right)}&{\left( {{z_3} – {z_1}} \right)} \end{array}} \right| = 0\)

As a result, the determinant form yields the Cartesian equation passing through three non-collinear points.

Solved Examples – Collinearity of Three Points

Q.1. Using the slope method, determine if the points \(A, B\), and \(C\) are collinear or not.

Ans: Let us consider \(A(1,2)=\left(x_{1}, y_{1}\right), B(2,3)=\left(x_{2}, y_{2}\right)\) and \(C(3,4)=\left(x_{3}, y_{3}\right)\)
We know that the slope of the line segment is slope \(=m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Now slope of line segment \(A B=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-2}{2-1}=1\)
And the slope of line segment \(B C=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}=\frac{4-3}{3-2}=1\)
Since the slope of line segment \(A B=\) slope of line segment \(B C\).
Therefore, the points \(A, B\), and \(C\) are collinear.

Q.2. Examine whether points \(P(-1,-1), Q(1,1)\), and \(R(3,3)\) are collinear.
Ans: Let us consider \(P(-1,-1)=\left(x_{1}, y_{1}\right), Q(1,1)=\left(x_{2}, y_{2}\right)\) and \(R(3,3)=\left(x_{3}, y_{3}\right)\).
We know that area of a triangle \(=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|\)
Now area of a triangle \(=\frac{1}{2}|(-1(1-3))+1(3-(-1))+3(-1-1)|\)
\(=\frac{1}{2}|(2+4-6)|\)
\(=\frac{1}{2}|0|=0\)
So, the area \(=0\)
Since the area of the given points is \(0\).
Therefore, the given points are collinear.

Q.3. The points \(A(8,1), B(15,7)\) and \(C(a, 3)\) are collinear. Find \(a\).
Ans: As given the points \(A, B\) and \(C\) are collinear we have
We know that the slope of the line segment is slope \(=m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
The slope of \(A B=\) slope of \(B C\)
\(\Rightarrow \frac{7-1}{15-8}=\frac{3-7}{a-15}\)
\(\Rightarrow \frac{6}{7}=\frac{-4}{a-15}\)
\(\Rightarrow 6(a-15)=-4 \times 7\)
\(\Rightarrow 6 a-90=-28\)
\(\Rightarrow 6 a=-28+90\)
\(\Rightarrow 6 a=62\)
\(\Rightarrow a=\frac{62}{6}=\frac{31}{3}\)
Hence, the value of \(a\) is \(\frac{31}{3}\).

Q.4. Find the vector form equation of the plane that passes through the points \((1,1,0)\), \((1,2,1)\), and \((-2,2,-1)\)
Ans: We’ll start by looking at the determinant of the three points to see if they’re collinear.
\(\left[ {\begin{array}{*{20}{c}} 1&1&0\\ 1&2&1\\ { – 2}&2&{ – 1} \end{array}} \right] = \, – 5\)
The fact that the determinant’s value is not zero indicates that the points are not collinear. Now we’ll find the Cartesian equation for the plane as follows:
\(\left| {\begin{array}{*{20}{c}} {\left( {x – 1} \right)}&{\left( {y – 1} \right)}&{\left( {z – 0} \right)}\\ {\left( {1 – 1} \right)}&{\left( {2 – 1} \right)}&{\left( {1 – 0} \right)}\\ {\left( { – 2 – 1} \right)}&{\left( {2 – 1} \right)}&{\left( { – 1 – 0} \right)} \end{array}} \right| = 0\)
\(\left| {\begin{array}{*{20}{c}} {\left( {x – 1} \right)}&{\left( {y – 1} \right)}&z\\ 0&1&1\\ { – 3}&1&{ – 1} \end{array}} \right| = 0\)
On solving, we get \((x-1)(-2)-(y-1)(3)+z(3)=0\)
\(\Rightarrow-2 x+2-3 y-3+3 z=0\)
\(\Rightarrow-2 x-3 y+3 z-1=0\)
\(2 x+3 y-3 z=-1\) is the obtained equation of the plane.

Q.5. If the given points are collinear \((8,1),(k,-4),(2,-5)\), then find the value of \(k\).
Ans: Let us consider \((8,1)=\left(x_{1}, y_{1}\right),(k,-4)=\left(x_{2}, y_{2}\right)\) and \((2,-5)=\left(x_{3}, y_{3}\right)\).
We know that area of a triangle \(=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|\)
Now area of a triangle \(=\frac{1}{2}|(8(-4-(-5)))+k(-5-1)+2(1-(-4))|=0\)
\(\Rightarrow \frac{1}{2}|(8-6 k+10)|=0\)
\(\Rightarrow \frac{1}{2}|18-6 k|=0\)
\(\rightarrow 18-6 k=0\)
\(\Rightarrow k=3\)
Hence, the value of \(k\) is \(3\).

Summary

In this article, we discussed the condition of collinearity of three points, check for collinearity of three points, the number of circles that can be drawn through three collinear points, the equation of the plane passing through three non – collinear points, solved examples on collinearity of three points, and FAQs on collinearity of three points.

The learning outcome of this article is we learned how to solve mathematical questions with conditions of collinearity of three points.

Frequently Asked Questions – Collinearity of Three Points

Q.1. What is the collinearity of a point?
Ans: Collinearity is the property of a set of points lying on a single line in geometry. Collinearity refers to a group of points that share this feature. The term has also been applied to aligned objects in a broader sense, such as those that are “in a line” or “in a row.”

Q.2. What are three non-collinear points?
Ans: Non-collinear points are three or more points that do not lie on the same straight line. If any of the points are not on the same line, they are considered non-collinear as a group.

Q.3. How do you find missing collinear points?
Ans: When three points are collinear, and one of them is missing a coordinate, we can use the area of the triangle idea to find the missing coordinate. If three points \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) are collinear, the area of triangle \(A B C\) equals zero.

Q.4. How many circles can pass through 3 collinear points?
Ans: There is no way to draw a circle through those collinear points. A straight line, on which the three points lie, is the only figure drawn through them. As a result, drawing a single circle through three collinear points is impossible.

Q.5. How to check if the three coordinates of points are collinear?
Ans: There are two approaches for determining whether the three points are collinear:
(i) The slope formula method and,
(ii) The area of triangle formula method.
Slope formula method: If the slope of any two pairs of points is the same, the three points are collinear.
With three points \(A, B\), and \(C\) three pairs of points can be formed. They are \(A C\).
If the slope of \(A B=\) slope of \(B C=\) slope of \(A C\), then \(A, B\), and \(C\) are collinear points.
Triangle formula method: If three distinct points \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) are collinear, then we cannot form a triangle because for such a triangle, there will be no altitude (height). Therefore, three points \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) will be collinear if the area of \(\triangle A B C=0\).

We hope this detailed article on the collinearity of three points helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!