Section Formula

The Section Formula is a useful tool in coordinate geometry that helps us find the coordinates of a point that divides a line segment in a known ratio. If a point divides a line segment into two parts equal or unequal, the section formula allows us to determine the coordinates of that dividing point.

The section formula can be applied in two ways:

• Internal Division : The point lies between the endpoints of the line segment and divides it internally.
• External Section Formula: The point lies outside the line segment, on its extension, and divides the line externally.

Because these two situations are different, we use two corresponding formulas:

1. Internal Section Formula
2. External Section Formula

Internal Section Formula

When the point divides the line segment in the ratio m: n internally at point C then that point lies on the line segment i.e., C divides AB internally, then we can use the Internal Section formula to calculate coordinates of C with the help of ratio and coordinates of the endpoints A and B. It is also called the Internal Division.

The term 'Sectional Formula' is generally used for Internal Sectional Formula. If the coordinates of A and B are (x₁, y₁) and (x₂, y₂) respectively then Internal Section Formula is given as:

Derivation of Internal Section Formula

Let  A(x₁, y₁) and B(x₂, y₂) be the endpoints of the given line segment AB and C(x, y) be the point that internally divides AB in the ratio m: n.

i.e., AC / CB = m / n

We want to find the coordinates of C (x, y). Take the given line segment and draw perpendiculars of A, C, and B parallel to Y coordinate joining at P, Q, and R on the X-axis as illustrated in the diagram below:

From above diagram,

AM = PQ = OQ - OP = (x - x₁),
CN = QR = OR - OQ = (x₂ -  x),
CM = CQ - MQ = (y - y₁),
and BN = BR - NR = (y₂ - y)

Clearly, ∆AMC ∼ ∆CNB (As all angles are equal for both triangles)
Therefore, ratio of sides are same for similar triangles.

⇒ AC / CB = AM / CN = CM / BN

Now substituting the values in the above relation

⇒ m / n = [(x - x₁)/(x₂ -x)] = [(y - y₁)/(y₂ - y)]
⇒ m / n = [(x - x₁)/(x₂ -x)]   and m / n = [(y - y₁)/(y₂ - y)]

Solving the 1^st condition,

⇒ m(x₂ - x) = n(x - x₁)
⇒  (m + n)x = (mx₂ + nx₁)
⇒  x = (mx₂ + nx₁) / (m + n)

Solving the 2^nd condition,

⇒  m(y₂ - y) = n(y - y₁)
⇒  (m + n)y = (my₂ + ny₁)
⇒  y = (my₂ + ny₁) / (m + n)

Therefore,

\bold{C (x, y) = \left( \frac{m x_2+ n x_1}{m + n} , \frac{m y_2 + n y_1}{m + n }\right)}

External Section Formula

When the point which divides the line segment in the ratio m : n lies outside the line segment i.e., when we extend the line it coincides with the point, then we can use the External Section formula to calculate the coordinates of C. It is also called External Division.

If the coordinates of A and B are (x₁, y₁) and (x₂, y₂) respectively then the external Section Formula is given as follows:

Derivation of the External Section Formula

To derive the internal section we took a line segment and a point C(x, y) which lies on the line, but in the case of the external section formula, we have to take that point C(x, y) outside the line segment.

Let A(x₁, y₁) and B(x₂, y₂)  be the endpoints of the given line segment AB and C(x, y) be the point that divides AB in the ratio m: n externally. To find the coordinates (x, y) of C, take the external divided line AB and draw perpendiculars from A, B, and C parallel to the y-axis joining at P, Q, and R on the x-axis.

Perpendicular from A and B parallel to the x-axis which meats the perpendicular drawn from C parallel to the y-axis at M and N. The following diagram shows the complete construction.

From the diagram,

AM = PR = OR - OP = (x - x₁),
BN = QR = OR - OQ = (x - x₂),
CM = RC - MR = (y - y₁),
and CN = CR - NR = (y - y₂)

Clearly, ∆AMC ∼ ∆CNB (As all angles are equal for both triangles)

As, the ratio of all the sides are same for similar triangles.
⇒ AC / BC = AM / BN = CM / CN

Now substituting the values in the above relation

⇒  m/n = [(x - x₁)/(x - x₂)] = [(y - y₁)/(y - y₂)]
⇒  m/n = [(x - x₁)/(x - x₂)] and m/n = [(y - y₁)/(y - y₂)]

Solving the 1st condition,

⇒  m(x - x₂) = n(x - x₁)
⇒  (m - n)x = (mx₂ - nx₁)
⇒  x = (mx₂ - nx₁)/(m - n)

Solving the 2nd condition,

⇒  m(y - y₂) = n(y - y₁)
⇒  (m - n)y = (my₂ - ny₁)
⇒  y = (my₂ - ny₁)/(m - n)

Therefore,

\bold{C(x,y) = \left(\frac{mx_2 - n x_1}{ m - n} ,  \frac{m y_2 - ny_1}{m - n }\right)}

Section Formula for Midpoint

When the point dividing the line segment joining the two points A and B coincide with the midpoint of the line segment, then this special case emerges. The section formula for this case is also referred to as the midpoint formula. 

Let two points A(x₁, y₁) and B(x₂, y₂), and a point C(x, y) be the midpoint of the line segment AB i.e., AC: CB = 1:1, then the section formula becomes as follows:

\bold{C(x, y) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)}

Related Articles:

• Distance Formula
• Point, Lines, and Planes
• Coordinate Geometry

Sample Problems of Section Formula

Problem 1: Find the coordinates of the midpoint of the line segment joining A(2, 5) and B(8, 1).

Solution: 

Let P(x, y) be the mid point of line segment joining A(2, 5) and B(8, 1).

Then, P(x, y) =( (x₁ + x₂)/2, (y₁ + y₂)/2)

⇒ P(x, y) = ( (2 + 8)/2,  (5 + 1)/2)
⇒ P(x, y) = (5, 3)

Therefore, the coordinates of the midpoint of the line segment AB are (5, 3).

Problem 2: Find the point P on the line joining A(2, 3) and B(5, 7) that is equidistant from points A and B.

Solution:

Let the coordinates of the point P be (x, y). 

PA = √{(x-2)² + (y-3)²)}
PB = √{(x-5)² + (y-7)²}

As P is equidistant from A and B, 

PA = PB
⇒ PA² = PB²
⇒ (x-2)² + (y-3)²= (x-5)² + (y-7)²
⇒ x²+4 - 4x + y² + 9 - 6y = x² + 25 - 10x + y² + 49 -14y
⇒ - 4x + 13 - 6y = 74 - 10x -14y
⇒ 6x + 8y = 61 . . .(1)

As line passing through two points (x₁, y₁) and (x₂, y₂) is given by:
y - y₁ = [(y₂ - y₁) / (x₂ - x₁)](x - x₁)

Thus, equation of  line joining A(2, 3) and B(5, 7) is

y - 3 = [(7-3)(5-2)(x - 2)]
⇒ y - 3 = (4/3)(x-2)
⇒ 3y - 9 = 4x - 8
⇒ 3y - 4x = 1
⇒ y = (1 + 4x)/3 . . .(2)

Using the section formula, we can find the coordinates of the point P that satisfies this equation. Letting m = 1 and n = 1, we have:

Px = (x1 + x2) / 2 = (2 + 5) / 2 = 3.5
Py = (y1 + y2) / 2 = (3 + 7) / 2 = 5

Therefore, the coordinates of the point P that is equidistant from A and B are (3.5, 5).

Problem 3: Find the coordinates of point C (x, y) where it divides the line segment joining (4, – 1) and (4, 3) in the ratio 3 1 internally. 

Solution: 

Given coordinates are A (4, -3) and B (8, 5)

Let C (x, y) be a point which divides the line segment in the ratio of 3 : 1  i.e m : n = 3 : 1 

Now using the formula C(x, y) = { (m × x2 + n × x1) / (m + n ) ,  (m × y2 + n × y1) / (m + n ) } as C is dividing internally.

⇒ C(x, y) = {(3×4 + 1×4 )/(3+1), (3×3  + 1×(-1))/(3+1)}
⇒ C(x, y) = {16/4, 8/4}
⇒ C(x, y) = {4, 2} 

Hence, the coordinates are (4, 2).

Problem 4: A(2, 7) and B(–4, –8) are the coordinates of the line segment AB. Two points intersected the segment. Find their coordinates of them.

Solution: 

Let S(x₁, y₁) and T(x₂, y₂) be the two points, which divides the line segment AB into 3 equal parts i.e.,

AS = ST = TB  . . . (1)

⇒  S divides the line segment AB in the ratio = AS/SB
⇒  Required ratio = AS/(ST + TB)
⇒  Required ratio = AS/(AS + AS)    [from equation (1)]
⇒  Required ratio = AS/2 AS
⇒  Required ratio =1/2

So, S divides the line segment AB in the ratio of 1 : 2

Now applying section formula to find the coordinates of point S

⇒  x₁ = (1 × (-4) + 2 × 2)/(1 + 2)
⇒  x₁ = (-4 + 4)/3
⇒  x₁ = 0

Similarly, for y coordinate,

⇒  y₁ =  (1 × (-8) + 2 × 7)/(1 + 2)
⇒  y₁ = (14 - 8)/3
⇒  y₁ = 2

Thus, S(x₁, y₁) = (0, 2) 

Similarly T divides line segment AB in 2:1.

Now applying section formula to find the coordinates of point T

⇒  x₂ = (2 × (-4) + 1 × 2)/(2 + 1)
⇒  x₂ = (-8 + 2)/3
⇒  x₂ = -2

Similarly, for y coordinate

⇒  y₂ =  (2 × (-8) + 1 × 7)/(2 + 1)
⇒  y₂ = (-16 + 7)/3
⇒  y₂ = -3

Thus, T (x₂, y₂) = (-2, -3)

Therefore the coordinates of required points are S (x₁, y₁) = (0, 2) and T (x₂, y₂) = (-2, -3).

Problem 5: A (4, 5) and B (7, -1) are two given points, and point C divides the line-segment AB externally in the ratio 4 : 3. Find the coordinates of C.

Solution:

Given coordinates are A (4, 5) and B (7, -1)

Let C (x, y) be a point which divides the line segment externally in the ratio of 4 : 3  i.e m : n = 4 : 3 

Now using the formula C(x, y) = { (m × x2 - n × x1) / (m - n) ,  (m × y2 - n × y1) / (m - n ) } as C is dividing internally.

C(x) = (mx₂ - nx₁)/(m - n)
⇒ C(x) = (4 × 7 - 3 × 4)/(4 - 3)
⇒ C(x) = (28 - 12)/(1)
⇒ C(x) = 16

C(y) = (my2 - ny1) / (m - n )
⇒ C(y) = (4 × (-1) - 3 × 5)/(4 - 3)
⇒ C(y) = (-4  - 15)/(1)
⇒ C(y) = -19

Hence, the required coordinates are (16, -19).

Problem 6: If a point P(k, 7) divides the line segment joining A(8, 9) and B(1, 2) in a ratio m :Two points then find values of m and n.

Solution: 

It is not mentioned that the point is dividing the line segment internally or externally. So, at that time we will consider the internal section as the default.

Given coordinates are A (8, 9) and B (1, 2)

Let the given point P (k, 7) divides the line segment in the ratio of m : 1

Using section formula for y coordinate.

⇒  7 =  (m y₂ + n y₁)/(m + n )
⇒  7 = (m × 2 + 1 × 9)/(m + 1)
⇒  7 = (2m + 9)/(m +1)
⇒  7m + 7 = 2m +9
⇒  5m = 2
⇒  m = 5 / 2

So the required ratio is 5 : 2
Therefore, value of m is 5 and value of n is 2

Problem 7: Line 2x+y−4=0 divides the line segment joining the points A(2,−2) and B(3,7). Find the ratio of a line segment in which the line is dividing.

Solution:

Given coordinates are A (2, -2) and B (3, 7).

Let the line with equation 2x + y - 4 = 0 divides the line segment at point C (x, y)
As (x, y) lies on the the 2x + y - 4 = 0, so we can use same variable for this instance only.

Let us assume the given line cuts the line segment in the ratio 1 : n.
By section formula,

⇒  x = (m x₂ + n x₁)/(m + n)
⇒  x = (3 + 2n)/(1 + n)  . . . (1)

Similarly,

⇒  y = (m y₂ + n y₁)/(m + n)
⇒  y = (7 - 2n)/(1 + n)  . . . (2)

Now substituting the equations (1) and (2) in the given equation of the line.

⇒  2x + y - 4 = 0
⇒  2 [(3 + 2n)/(1 + n)] +  [(7 - 2n)/(1 + n)] - 4 = 0  
⇒  6 + 4n + 7 − 2n − 4(1 + n) = 0
⇒ 13 + 2n − 4 − 4n = 0
⇒ 9 − 2n = 0
⇒  n = 2 / 9

Therefore the ratio at which the line divides is 9 : 2. 

Note: We can also find the values of x and y by substituting the value of n in the equation (1) and (2).

Practice Questions on Section Formula

Question 1: Calculate the midpoint of the line segment joining C(1,−3) and D(4,5).

Question 2: Find the coordinates of the point that divides the line segment joining A(2, 3) and B(10,7) in the ratio 3:1 internally.

Question 3: The points A(2,−2) and B(8,4) are trisected by points P and Q. Find the coordinates of P and Q.

Question 4: The point P(2,4) divides the line segment joining Q(1,2) and R(4,8) in the ratio m:n. Determine the values of m and n.

Question 5: Determine the coordinates of the point that divides the line segment joining A(−4,2) and B(6,8) in the ratio 5:3 externally.