ICSE Class 10 Maths Chapter 13 Section and Mid Point Formula

Section And Mid-Point Formula

13.1 Introduction

For any two known (given) points in a co-ordinate (Cartesian) plane, the knowledge of co-ordinate geometry may be used to find:

(i) the distance between the given points,

(ii) the co-ordinates of a point which divides the line joining the given points in a given ratio,

(iii) the co-ordinates of the mid-point of the line segment joining the two given points,

(iv) equation of the straight line through the given points,

(v) equation of the perpendicular bisector of the line segment obtained on joining the given two points, etc.

13.2 The Section Formula

To find the co-ordinates of a point which divides the line segment joining two given points in a given ratio.

(If a point P lies in a line segment joining the points A and B, then P divides AB in the ratio AP : PB).

Let AB be a line joining the points A = (x₁, y₁) and B = (x₂, y₂) and point P divides the line segment AB in the ratio m₁ : m₂.

i.e. $\frac{\text{AP}}{\text{PB}}=\frac{m_1}{m_2}$

Required to find: The co-ordinates of point P.

Let P = (x, y)

Draw AL, PM and BN perpendiculars on the x-axis. Thus, AL, PM and BN are parallel lines. It is clear from the figure that:

AR = LM = OM - OL = x - x₁;

PR = PM - RM = PM - AL = y - y₁;

PS = MN = ON - OM = x₂ - x

and, BS = BN - SN = BN - PM = y₂ - y

Since, Triangle APR and Triangle PBS are similar.

$\frac{\text{AR}}{\text{PS}}=\frac{\text{PR}}{\text{BS}}=\frac{\text{AP}}{\text{PB}}$ [Corresponding sides of similar triangles are in proportion]

$\frac{\text{AR}}{\text{PS}}=\frac{\text{AP}}{\text{PB}}=>\frac{x}{-}=\frac{m_1}{m_2}$

$=>m_{2}x-m_2{x}_1=m_1{x}_2-m_{1}x$ [By cross multiplication]

$=>m_{1}x+m_{2}x=m_1{x}_2+m_2{x}_1$

$=>x(m_1+m_2)=m_1{x}_2+m_2{x}_1$

$\therefore x=\frac{m_1}{x_2}$

Since,

$\frac{\text{PR}}{\text{BS}}=\frac{\text{AP}}{\text{PB}}=>\frac{y}{-}=\frac{m_1}{m_2}=>y=\frac{m_1}{y_2}$

Co-ordinates of P = $(\frac{m_1}{x_2},\frac{m_1}{y_2})$

Problem 1

Find the co-ordinates of point P which divides the join of A (4, -5) and B (6, 3) in the ratio 2 : 5.

Solution:

Let the co-ordinates of P be (x, y)

$\therefore x=\frac{m_1}{x_2}=\frac{2}{\times }=\frac{32}{7}$

and, $y=\frac{m_1}{y_2}=\frac{2}{\times }=\frac{-}{19}$

$\therefore P=(\frac{32}{7},\frac{-}{19})$

Answer: P = (32/7, -19/7)

Teacher's Note

The section formula helps us find positions of objects dividing a path, similar to finding a meeting point on a straight road between two cities.

Problem 2

Find the ratio in which the point (5, 4) divides the line joining points (2, 1) and (7, 6).

Solution:

Let the required ratio be m₁ : m₂. Take (2, 1) = (x₁, y₁), (7, 6) = (x₂, y₂) and (5, 4) = (x, y)

$x=\frac{m_1}{x_2}=>5=\frac{m_1}{\times }$

$=>5m_1+5m_2=7m_1+2m_2$

$=>2m_1=3m_2$

$=>\frac{m_1}{m_2}=\frac{3}{2}$

The required ratio is 3 : 2.

Answer: 3 : 2

Teacher's Note

Reverse calculations in geometry help us verify positions, just as checking a receipt confirms a purchase location.

Problem 3

In what ratio is the line joining the points (4, 2) and (3, -5) divided by the x-axis? Also, find the co-ordinates of the point of intersection.

Solution:

Let the required ratio be k : 1 and the point on the x-axis be (x, 0).

Since, $y=\frac{k_y}{+}$ [Taking (4, 2) = (x₁, y₁) and (3, -5) = (x₂, y₂)]

$=>0=\frac{k}{\times }$

$=>0=-5k+2$

$=>k=\frac{2}{5}$

$=>m_1:m_2=2:5$

Now, $x=\frac{2}{\times }=\frac{26}{7}$

The ratio = 2 : 5 and the required point of intersection = (26/7, 0)

Answer: The ratio is 2 : 5 and the required point of intersection = (26/7, 0)

Teacher's Note

Finding where a path crosses an axis is like determining when a plane crosses the equator on its journey between destinations.

This is a preview of the first 3 pages. To get the complete book, click below.