Sample Problems On Section Formula

Problem 1: 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 2: If a point P(k, 7) divides the line segment joining A(8, 9) and B(1, 2) in a ratio m : n 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 3: 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 4: 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 5: 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 varible 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).

Problem 6: 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 7: A(2, 7) and B(–4, –8) are the coordinates of the line segment AB. There are two points that trisected the segment. Find their coordinates of them.

Solution: 

TLet 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 segement 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).

Section Formula is a useful tool in coordinate geometry, which helps us find the coordinate of any point on a line which is dividing the line into some known ratio. Suppose a point divides a line segment into two parts which may be equal or not, with the help of the section formula we can find the coordinates of that point.

In this article, we will learn about section formulas, the types of division of lines, and how to solve problems based on them in detail.