Section Formula
Section Formulae
Simply, section formulae refer to the external and internal division of a line segment by a given point.
Section formulae have two types. They are,
 Section formulae for an internal division.
 Section formulae for an external division.
Section Formulae for Internal Division
Let's take a line with two ends point A(x_{1}, y_{2}) and B(x_{2}, y_{2}) which are joined by the line segment AB. Consider P(x, y) be any point on AB which divides the line internally in the ratio m_{1}:m_{2}
i.e. AP:PB = m_{1}:m_{2}
The formula for the section formulae in internal division is (x, y) = (\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\)),(\(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\))
Section Formulae for External Division
If the point P(x, y) divides AB externally in the ratio of m_{1}:m_{2} then the divided segment BP is measured in opposite direction and hence m_{2} is taken as negative.
\(\therefore\) The section formulae for external division is,
(x, y) = (\(\frac{m_1x_2  m_2x_1}{m_1  m_2}\)),(\(\frac{m_1y_2  m_2y_1}{m_1  m_2}\))
In special case, the midpoint formulae is also used'
m_{1}:m_{2} = 1:1 i.e. m_{1} = m_{2}\(\therefore\) x = \(\frac{x_1 + x_2}{2}\) and y = \(\frac{y_1 + y_2}{2}\)
Thus, coordinates P(x, y) are P(\(\frac{x_1 + x_2}{2}\),\(\frac{y_1 + y_2}{2}\)) which is called midpoint formulae.
 Internal division(x,y) = (\(\frac{m_1x_1+m_2x_1}{m_1+m_2}\)), (\(\frac{m_1y_1+m_2y_1}{m_1+m_2}\))
 External division(x,y) = (\(\frac{m_1x_1m_2x_1}{m_1m_2}\)), (\(\frac{m_1y_1m_2y_1}{m_1m_2}\))
 Midpoint formulae = (\(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\))
Solution:
Here,
(4, 6) = (x_{1}, y_{1})
(2, 4) = (x_{2}, y_{2})
The coordinates of the mid point = (\(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\))
= (\(\frac{4+ 2}{2}\), \(\frac{6 + 4}{2}\))
= (\(\frac{6}{2}\), \(\frac{10}{2}\))
= (3, 5)
\(\therefore\) The coordinates of given mid point is (3, 5).
Solution:
Here,
(2, 3) = (x_{1}, y_{1})
(7, 3) = (x_{2}, y_{2})
m_{1} : m_{2} = 3 : 2
Now,
or, (x, y) = (\(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\), \(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\))
or, (x, y) = (\(\frac{3\times7 + 2\times2}{3 + 2}\), \(\frac{3\times3 + 2\times3}{3 + 2}\))
or, (x, y) = (\(\frac{21 + 4}{5}\), \(\frac{9 + 6}{5}\))
or, (x, y) = (\(\frac{25}{5}\), \(\frac{15}{5}\))
or, (x, y) = (5, 3)
So, the required point is (5, 3).
Solution:
Let the points on yaxis be (0, y) which divides the line joining the points
(4, 1) = (x_{1}, y_{1}) and (10, 1) = (x_{2}, y_{2})
m_{1} : m_{2} = ?
By using sectional formula
or, x = \(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\)
or, 0 =\(\frac{m_1 \times10+ m_2 \times(4)}{m_1 + m_2}\)
or, 0 = 10m_{1 } 4m_{2}
or, \(\frac{m_1}{m_2}\) = \(\frac{4}{10}\)
or, \(\frac{m_1}{m_2}\) = \(\frac{2}{5}\)
\(\therefore\) m_{1} : m_{2} = 2 : 5
Solution:
Let the points on yaxis be (x, 0) which divides the line joining the point
(3, 2) = (x_{1}, y_{1})
(3,9) = (x_{2}, y_{2})
m_{1} : m_{2}= ?
By using section formula,
or, y = \(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\)
or, 0 =\(\frac{m_1\times(9)+ m_2 \times2}{m_1 + m_2}\)
or, 0 = 9m_{1}+ 2m_{2}
or, \(\frac{m_1}{m_2}\) = \(\frac{2}{9}\)
\(\therefore\)m_{1} : m_{2}= 2 : 9
Solution:
Given,
(5a, 7b) = (x_{1}, y_{1})
(3a,2b) = (x_{2}, y_{2})
Coordinates of mid point = (\(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\))
= (\(\frac{5a + 3a}{2}\), \(\frac{7b 2b}{2}\))
= (\(\frac{8a}{2}\), \(\frac{5b}{2}\))
= (4a, \(\frac{5b}{2}\))
Solution:
(2,4) = (x_{1}, y_{1})
(3, 6) = (x_{2}, y_{2})
m_{1} : m_{2} = 2 : 3
Now,
or, (x, y) = (\(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\), \(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\))
or, (x, y) = (\(\frac{2\times (3) + 3\times2}{2 +3}\), \(\frac{2\times6 + 3\times (4)}{2 + 3}\))
or,(x, y) = (\(\frac{ 6+ 6}{5}\), \(\frac{1212}{5}\))
or, (x, y) = (0, 0)
\(\therefore\)The required point is (0, 0).
Solution:
(10, 12) = (x_{1}, y_{1})
(3, 9) = (x_{2}, y_{2})
m_{1} ; m_{2} = 4 : 3
Now,
or, (x, y) = (\(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\), \(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\))
or, (x, y) = (\(\frac{4 \times(3) + 3 \times(10)}{4 + 3}\), \(\frac{4 \times(9) + 3 \times12}{4 + 3}\))
or, (x, y) = (\(\frac{12  30}{7}\), \(\frac{36 + 36}{7}\))
or, (x, y) = (6, 0)
\(\therefore\) The required points are (6, 0)
Solution:
(8, 5) = (x_{1}, y_{1})
(12, 7) = (x_{2}, y_{2})
Coordinates of the mid point = (\(\frac{x_1 + x_2}{2}\), \(\frac{x_1 + y_2}{2}\))
= (\(\frac{8  12}{2}\), \(\frac{5  7}{2}\))
= (\(\frac{ 4}{2}\), \(\frac{ 2}{2}\))
= (2, 1)
Solution:
Given,
(3, 7) = (x_{1}, y_{1})
(5, 3) = (x_{2}, y_{2})
Coordinates of the mid point = (\(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\))
= (\(\frac{3 + (5)}{2}\), \(\frac{7 + (3)}{2}\))
= (\(\frac{3  5}{2}\), \(\frac{7  3}{2}\))
= (\(\frac{8}{2}\), \(\frac{10}{2}\))
= (4, 5)
Solution:
(2, 2) = (x_{1}, y_{1})
(2, 2) = (x_{2}, y_{2})
m_{1} : m_{2} = 2 : 5
now,
or, (x, y) = (\(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\), \(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\))
or, (x, y) = (\(\frac{2 \times\ (2) + 5 \times\ 2}{2 + 5}\), \(\frac{2 \times\ (2) + 5 \times\ 2}{2 + 5}\))
or, (x, y) = (\(\frac{4 + 10}{7}\), \(\frac{4 + 10}{7}\))
or, (x, y) = (\(\frac{6}{7}\), \(\frac{6}{7}\))
\(\therefore\) The required points is (\(\frac{6}{7}\), \(\frac{6}{7}\))
Solution:
(0, 3) = (x, y)
(6, 8) = (x_{1}, y_{1})
Let coordinates of other end be (x_{2}, y_{2})
Now,
or, x = \(\frac{x_1 + x_2}{2}\)
or, 0 = \(\frac{6 + x_2}{2}\)
or, 0 =  6 + x_{2}
\(\therefore\) x_{2 = }6_{}Again,
or, y = \(\frac{y_1 + y_2}{2}\)
or, 3 = \(\frac{8 + y_2}{2}\)
or, 6 = 8 + y_{2}
or, 6  8 = y_{2}
\(\therefore\) y_{2}= 2
So, the coordinates of other end is (6, 2).
Solution:
Given,
(2, 2) = (x_{1}, y_{1})
(2,2) = (x_{2}, y_{2})
(x, y) = (\(\frac{6}{7}\), \(\frac{6}{7}\))
m_{1} : m_{2} = ?
By using section formula,
or, x = \(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\)
or, \(\frac{6}{7}\) =\(\frac{m_1 \times( 2) + m_2 \times 2}{m_1 + m_2}\)
or, 6(m_{1}+ m_{2}) = 7(2m_{1} + 2m_{2})
or, 6m_{1}+ 6m_{2} = 14m_{1} + 14m_{2}or, 6m_{1}+ 14m_{1}=14m_{2 }6m_{2}
or, 20m_{1} = 8m_{2}
or, \(\frac{m_1}{m_2}\) = \(\frac{8}{20}\)
\(\therefore\) m_{1} : m_{2}= 2 : 5
Now,
or, y = \(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\)
or, \(\frac{6}{7}\) = \(\frac{2\times (2)+ 5\times2}{2 + 5}\)
or, \(\frac{6}{7}\) = \(\frac{(4)+ 10}{7}\)
or, \(\frac{6}{7}\) = \(\frac{6}{7}\)
\(\therefore\) The required ratio is 2 : 5.
Solution:
(4, 5) = (x, y)
(5, 3) = (x_{1}, y_{1})
Let the coordinates of the other end be (x_{2}, y_{2})
or, x = \(\frac{x_1 + x_2}{2}\)
or, 4 = \(\frac{5 + x_2}{2}\)
or, 8 = 5 + x_{2}
or, 8 + 5 = x_{2}
or, x_{2} = 3
Again,
or, y = \(\frac{y_1 + y_2}{2}\)
or, 5 = \(\frac{3 + y_2}{2}\)
or, 10 = 3 + y_{2}
or, y_{2} = 10 + 3
or, y_{2} = 7
\(\therefore\) The required point(x_{2}, y_{2}) = (3, 7)
Solution:
(2, 4) = (x1, y1)
(3, 6) = (x2, y2)
(0, 0) = (x, y)
By using section formula
or, x = \(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\)
or, 0 = \(\frac{m_1 \times\ (3) + m_2 \times 2}{m_1 + m_2}\)
or, 0 = \(\frac{3m_1 + 2m_2}{m_1 + m_2}\)
or, 0 = 3m_{1} + 2m_{2}
or, 3m_{1} = 2m_{2}
or, \(\frac{m_1}{m_2}\) = \(\frac{2}{3}\)
\(\therefore\) m_{1} : m_{2} = 2 : 3
Now,
or, y = \(\frac{m_1 y_2 + m_2 y_1}{2 + 3}\)
or, 0 = \(\frac{2 \times\ 6 + 2\times\ (4)}{m_1 + m_2}\)
or, 0 = \(\frac{12  12}{5}\)
or, 0 = 0
\(\therefore\) The required ratio is 2 : 3
Solution:
(1, 3) = (x_{1}, y_{1})
(4, 8) = (x_{2}, y_{2})
(2, 6) = (x, y)
By using section formula
or, x = \(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\)
or, 2 = \(\frac{m_1 \times\ 4 + m_2 \times\ (1)}{m_1 + m_2}\)
or, 2(m_{1} + m_{2}) = 4m_{1}  1m_{2}
or, 2m_{1} + 2m_{2} = 4m_{1}  1m_{2}
or, 2m_{1} 4m_{1} = 1m_{2}  2m_{2}
or, 2m_{1} = 3m_{2}
Now,
or, y = \(\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\)
or, 6 = \(\frac{3\times8 + (2)\times\ 3}{3 + (2)}\)
or, 6 = \(\frac{24 6}{3  2}\)
or, 6 = \(\frac{30}{5}\)
or, 30 = 30
\(\therefore\) The required ratio is (3, 2)

Find the coordinates of the mid point of the line segement joining the given points, (4, 6) and (2, 4).
(3, 5)
(3, 3)
(5, 5)
(3, 2)

Find the coordinates of the point (2, 3) and (7, 3) which divides the line segment joining the following points in the 3:2 ratio.
3, 5
5, 3
4, 1
2, 3

Find the ratio in which the yaxis divides the line segment joining the points (−4, 1) and (10, 1)
2 : 5
2 : 3
3 : 5
6 : 4

Find the ratio in which xaxis divides the line segment joining the points (3, 2) and (3, −9).
3 : 5
2 : 9
3 : 9
2 : 3

Find the coordinates of the mid point of the line segment joining the given points (5a, 7b) and (3a, −2b).
(8, (frac{5b}{2}))
(3, (frac{5b}{3}))
(4, (frac{6b}{4}))
(4, (frac{5b}{2}))

Find the coordinates of the point which divides the line segment joining the (2, −4) and (−3, 6) points in the 2:3 ratio.
0, 0
2, 1
2, 3
1, 2

Find the coordinates of the point which divides the line segment joining (10, 12) and (3, 9) points in the 4:3 ratio.
0, 6
6, 0
1, 6
6, 0

Find the coordinates of the mid point of the line segment joining the (8, 5) and (12, 7) points.
2, 1
2, 2,
1, 2
3, 1

Find the coordinates of the mid point of the line segment joining (3, 7) and (5, 3) points.
4, 5
4, 5
5, 4
4, 5

Find the coordinates of the point which divides the line segment joining (2, 2) and (2, 2) points in the 2:5 ratios.
2 ,6
5, 6
((frac{5}{7}), (frac{6}{5}))
((frac{6}{7}), (frac{6}{7}))

Find coordinates: (1, 3) and (4, 6), ratio 2:1
1, 3
2, 3
1, 2
3, 3

Find coordinates: (3, 7) and (5, 3)
5, 4
4, 5
4, 5
4, 5

Find coordinates: (6, 0) and (0, 8)
3, 4
3, 4
3, 4
4, 3

Find coordinates: (6, 8) and (6, 2)
1, 2
0, 3
3, 0
3, 0

Find coordinates: (4, 6) and (2, 4)
5, 3
3, 2
3, 5
2, 3

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Jan 15, 2017 
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pankajShow that the points A (3,4) B(7,3) and C (4,8) are the vertices of an isosceles 
Jan 01, 2017 
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