Collinearity of Three Points, Section Formula and Areas of Triangles, Class 10, Science | Extra Documents, Videos & Tests for Class 10 PDF Download

COLLINEARITY OF THREE POINTS

Let A, B and C three given points. Point A, B and C will be collinear, if the sum of lengths of any two linesegments is equal to the length of the third line-segment.

In the adjoining fig. there are three point A, B and C.

Three points A, B and C are collinear if and only if
(i) AB + BC = AC
or (ii) AB + AC = BC
or (iii) AC + BC = AB

SECTION FORMULA

Coordinates of the point, dividing the line-segment joining the points (x₁, y₁ ) and (x₂, y₂) internally in the ratio m₁ : m₂ are given by

Proof. Let P (x, y) be the point dividing the line-segment joining A (x₁, y₁ ) and B (x₂, y₂) internally in the ratio
m₁ : m₂. We draw the perpendiculars AL, BM and PQ on the x-axis from the points A, B and P respectively.
L, M and Q are the points on the x-axis where these perpendiculars meet the x-axis.

We draw AC ⊥ PQ and PD ⊥  BM. Here AC || x-axis and PD || x-axis.

AC = LQ = OQ – OL = (x - x₁)
PD = QM = OM – OQ = (x₂ – x)
Putting in (1), we get

 m₂x – m₂x₁ = m₁x₂ – m₁x
 m₁x + m₂x = m₁x₂ + m₂x₁  ⇒ (m₁ + m₂)x = m₁x₂ + m₂x₁

Now, PC = PQ – CQ = PQ – AL = (y – y₁)
BD = BM – DM = BM – PQ = (y₂ – y)

Putting in (2), we get

Therefore, the coordinates of the point P are

Remark : To remember the section formula, the diagram given below is helpful:

Point Dividing a Line Segment in the Ratio k : 1 

if  :divides the line-segment, joining A (x₁, y₁) and B (x₂, y₂) internally in the ratio m₁ : m₂ we can express it as below :

 {By dividing the numerator and the denominator by m₂]

Putting  the ratio becomes k : 1 and the coordinates of P are expressed in the form

Therefore, the coordinates of the point P, which divides the line-segment joining A (x₁, y₁) and B (x₂, y₂) internally in the ratio k : 1, are given by 

Mid-point Formula : Coordinates of the mid-point of the line-segment joining (x₁, y₁) and (x₂, y₂) are   The mid-point M (x, y) of the line-segment joining A (x₁, y₁) and B(x2, y₂) divides the line-segment AB in the ratio 1 : 1. Putting m₁ = m₂ = 1 in the section formula, we get the coordinates of the mid-point as

Collinearity of three points :

Three given points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are said to be collinear if one of them must divide the line segment joining the other two points in the same ratio.

Remark : Three points are called non-collinear if one of them divides the line segment joining the other two points in different ratios

Ex.5 Find the co-ordinates of the points which divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts.

It is given that AB is divided into four equal parts : AP = PQ = QR = RB

Q is the mid-point of AB, then co-ordinates of Q are :

P is the mid-point of AQ, then co-ordinates of P are:

Also, R is the mid-point of QB, then co-ordinates of R are:

Hence, required co-ordinates of the points are:

Ex.6  If the point C(–1,2) divides the lines segment AB in the ratio 3 : 4, where the co-ordinates of A are (2, 5), find the coordinates of B.
 Sol. Let C (–1, 2) divides the line joining A (2, 5) and B (x, y) in the ratio 3 : 4. Then,

The coordinates of B are : B (–5, – 2)

Ex.6 Find the ratio in which the line segment joining the points (1, – 7) and (6, 4) is divided by x-axis.
Sol. Let C (x, 0) divides AB in the ratio k : 1.
By section formula, the coordinates of C are given by :

But C (x, 0) = 

Ex.7  Find the value of m for which coordinates (3,5), (m,6) and are collinear

Sol. Let P (m, 6) divides the line segment AB joining A (3,5), Bin the ratio k : 1.

Applying section formula, we get the co-ordinates of P :

 15k + 10 = 12(k + 1)
15k + 10 = 12k + 12

 15k – 12k = 12 – 10  
 3k = 2

Ex.8 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the co-ordinates of the other two vertices.
 Sol. Let ABCD be a square and two opposite vertices of it are A(–1, 2) and C(3, 2). ABCD is a square.
 AB = BC
 AB² = BC²
 (x + 1)² + (y – 2)2 = (x – 3)² + (y – 2)²
 x² + 2x + 1 = x₂ – 6x + 9
 2x + 6x = 9 – 1 = 8
 8x = 8 x = 1
ABC is right Δ at B, then
AC² = AB²+ BC²(Pythagoras theorem)

(3 + 1)² + (2 – 2)² = (x + 1)² + (y – 2)² + (x – 3)² + (y – 2)²

16 = 2(y – 2)² + (1 + 1)² + (1 – 3)²

16 = 2(y – 2)² + 4 + 4 2(y – 2)² = 16 – 8 = 8

(y – 2)² = 4  y – 2 =  2  y = 4 and 0

i.e. when x = 1 then y = 4 and 0
Co-ordinates of the opposite vertices are : B(1, 0) or D(1, 4)

AREA OF A TRIANGLE

In your previous classes, you have learnt to find the area of a triangle in terms of its base and corresponding altitude as below:

In case, we know the lengths of the three sides of a triangle, then the area of the triangle can be obtained by using the Heron's formula. In this section, we will find the area of a triangle when the coordinates of its three vertices are given. The lengths of the three sides can be obtained by using distance formula but we will not prefer the use of Heron's formula.

Some times, the lengths of the sides are obtained as irrational numbers and the application of Heron's formula becomes tedious. Let us develop some easier way to find the area of a triangle when the coordinates of its vertices are given.
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the given three points. Through A draw AQ ⊥OX, through B draw

BP ⊥ OX and through C draw CR ⊥ OX.

From the fig. AQ = y₁, BP = y₂ and CR = y₃, OP = x₂, OQ = x₁ and OR = x₃

_{PQ =} X₁– X_{2 ; QR =} X₃ – X_{1 and PR =} X₃–-X₂

Condition of collinearity of three points :
The given points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) will be collinear if the area of the triangle formed by them must be zero because triangle can not be formed.

is the required condition for three points to be collinear.

Ex.9 The co-ordinates of the vertices of ΔABC are A(4, 1), B(–3, 2) and C(0, k). Given that the area of ΔABC is 12 unit². Find the value of k.
Sol. Area of ΔABC formed by the given-points A(4, 1), B(–3, 2) and C(0, k) is

But area of ΔABC = 12 unit² ............ (given)

Ex.10  Find the value of p for which the points (–1, 3), (2, p), (5, –1) are collinear.
Sol. The given points A (–1, 3), B (2, p), C (5, – 1) are collinear.

 Area ΔABC formed by these points should be zero.
 The area of ΔABC = 0

Hence the value of p is 1.