# Division of Line Segment in Given Ratio – Constructions | Class 10 Maths

**Line** is a straight one-dimensional figure that has no thickness. In geometry, a line extends endlessly in both directions. It is described as the shortest distance between any two points. A-line can also be understood as multiple points connected to each other in one specific direction without a gap between them.

**Line Segment** is a part of a line that is bounded by two different endpoints and contains every point on the line between its endpoints in the shortest possible distance.

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### Rules to write down the coordinates of the point which divides the join of two given points P(x_{1},y_{1}) and Q(x_{2},y_{2}) internally in a given ratio m_{1}:m_{2}

- Draw the line segment joining the given points P and Q
- Write down the coordinates of p and Q at extremities
- Let R(x, y) be the input which divides PQ internally in the ratio m
_{1}:m_{2} - For x-coordinate of R, multiply x
_{2 }with m_{1}and x_{1}with m_{2 }and add the products. Divide the sum by m_{1}+m_{2} - For y-coordinate of R, multiply y
_{2}with m_{1 }and y_{1}with m_{2}and add the products. Divide the sum by m_{1}+m_{2}

### Derivation of Formula

Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be the two ends of a given line in a coordinate plane, and R(x,y) be the point on that line which divides PQ in the ratio m_{1}:m_{2 }such that

PR/RQ=m_{1}/m_{2} …(1)

Drawing lines PM, QN, and RL perpendicular on the x-axis and through R draw a straight line parallel to the x-axis to meet MP at S and NQ at T.

Hence, from the figure, we can say,

SR = ML = OL – OM = x – x

_{1}…(2)RT = LN = ON – Ol = x

_{2 }– x …(3)PS = MS – MP = LR – MP = y – y

_{1}…(4)TQ = NQ – NT = NQ – LR = y

_{2 }– y …(5)

Now ∆SPR is similar to ∆TQR

SR / RT = PR / RQ

By using equation 2, 3, and 1, we know,

x – x

_{1}/ x_{2 }– x = m_{1 }/ m_{2}m

_{2}x – m_{2}x_{1}= m_{1}x_{2}– m_{1}xm

_{1}x + m_{2}x = m_{1}x_{2}+ m_{2}x_{1}(m

_{1}+ m_{2})x = m_{1}x_{2 }+ m_{2}x_{1}x = (m

_{1}x_{2 }+ m_{2}x_{1}) / (m_{1}+ m_{2})

Now ∆SPR is similar to ∆TQR,

Therefore,

PS/TQ = PR/RQ

By using equation 4, 5, and 1, we know,

y – y

_{1}/ y_{2 }– y = m_{1}/m_{2}m

_{2}y – m_{2}y_{1}= m_{1}y_{2}– m_{1}ym

_{1}y + m_{2}y = m_{1}y_{2}+ m_{2}y_{1}(m

_{1}+ m_{2})y = m_{1}y_{2 }+ m_{2}y_{1}y = (m

_{1}y_{2 }+ m_{2}y_{1})/(m_{1}+ m_{2})

Hence, the coordinate of R(x,y) are,

### Sample Problems based on the Formula

**Problem 1: Calculate the co-ordinates of the point P which divides the line joining A(-3,3) and B(2,-7) in the ratio 2:3**

**Solution: **

Let (x, y) be th co-ordinates of the point P which divides the line joining A(-3, 3) and B(2, -7) in the ratio 2:3, then

Applying the formulae,

x = (m

_{1}x_{2 }+ m_{2}x_{1}) / (m_{1}+ m_{2})x = {2 x 2 + 3 x (-3)} / (2 + 3)

x = (4 – 9) / 5

x = -5 / 5

x = -1

y = (m

_{1}y_{2 }+ m_{2}y_{1}) / (m_{1}+ m_{2})y = {2 x (-7) + 3 x 3} / (2 + 3)

y = (-14 + 9) / 5

y = -5 / 5

y = -1

Hence the co-ordinates of point P are (-1, -1).

**Problem 2: If the line joining the points A(4,-5) and B(4,5) is divided by point P such that AP/AB=2/5, find the coordinates of P.**

**Solution:**

Given AP/AB = 2/5

5AP = 2AB = 2(AP + PB)

3AP = 2PB

AP/PB = 2/3

AP:PB = 2:3

Thus the point P divides the line segment joining the points A(4, -5) and B(4, 5) in the ratio 2:3 internally.

Hence the coordinates of P are,

P

_{x }= (2 x 4 + 3 x 4)/(2 + 3)= (8 + 12)/5

= 20/5

= 4

P

_{y }= (2 x 5 + 3 x (-5))/(2 + 3)= (10 – 15)/5

= -5/5

= -1

Coordinates of P are (4, -1).

**Problem 3: In what ratio does the point P(2, -5) divide the line segment joining the points A(-3, 5) and B(4, -9).**

**Solution:**

Let P(2, -5) divide the line segment joining the points A(-3, 3) and B(4, -9) in the ratio k:1, i.e. AP:PB = k:1

Therefore, Coordinates of P are,

P

_{x }= (k.4 + 1.(-3)) / (k + 1)P

_{y }= (k.(-9) + 1 x 5)/(k + 1)But P is (2, -5) therefore,

(4k – 3) / (k + 1) = 2 and (-9k + 5) / (k + 1) = -5

Solving any of the two equation we get k = 5/2

Therefore, the required ratio is 5/2:1 i.e. 5:2(internally)