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,

  1. Section formulae for an internal division.
  2. Section formulae for an external division.

Section Formulae for Internal Division
Let's take a line with two ends point A(x1, y2) and B(x2, y2) 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 m1:m2
i.e. AP:PB = m1:m2
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 formula for internal division

Section Formulae for External Division
If the point P(x, y) divides AB externally in the ratio of m1:m2 then the divided segment BP is measured in opposite direction and hence m2 is taken as negative.

.
Section formula for external division

\(\therefore\) 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'
m1:m2 = 1:1 i.e. m1 = m2
\(\therefore\) x = \(\frac{x_1 + x_2}{2}\) and y = \(\frac{y_1 + y_2}{2}\)
Thus, co-ordinates P(x, y) are P(\(\frac{x_1 + x_2}{2}\),\(\frac{y_1 + y_2}{2}\)) which is called mid-point 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) = (\(\frac{m_1x_1-m_2x_1}{m_1-m_2}\)), (\(\frac{m_1y_1-m_2y_1}{m_1-m_2}\))
  • Mid-point formulae = (\(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)) 

Solution:

Here,
(4, 6) = (x1, y1)
(2, 4) = (x2, y2)
The co-ordinates 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 co-ordinates of given mid point is (3, 5).

Solution:

Here,
(2, 3) = (x1, y1)
(7, 3) = (x2, y2)
m1 : m2 = 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 y-axis be (0, y) which divides the line joining the points
(-4, 1) = (x1, y1) and (10, 1) = (x2, y2)
m1 : m2 = ?
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 = 10m1 - 4m2
or, \(\frac{m_1}{m_2}\) = \(\frac{4}{10}\)
or, \(\frac{m_1}{m_2}\) = \(\frac{2}{5}\)
\(\therefore\) m1 : m2 = 2 : 5

Solution:

Let the points on y-axis be (x, 0) which divides the line joining the point
(3, 2) = (x1, y1)
(3,-9) = (x2, y2)
m1 : m2= ?
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 = -9m1+ 2m2
or, \(\frac{m_1}{m_2}\) = \(\frac{2}{9}\)
\(\therefore\)m1 : m2= 2 : 9

Solution:

Given,
(5a, 7b) = (x1, y1)
(3a,-2b) = (x2, y2)
Co-ordinates 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) = (x1, y1)
(-3, 6) = (x2, y2)
m1 : m2 = 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) = (\(\frac{- 6+ 6}{5}\), \(\frac{12-12}{5}\))
or, (x, y) = (0, 0)
\(\therefore\)The required point is (0, 0).

Solution:

(-10, 12) = (x1, y1)
(-3, -9) = (x2, y2)
m1 ; m2 = 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) = (x1, y1)
(-12, -7) = (x2, y2)

Co-ordinates 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) = (x1, y1)
(-5, -3) = (x2, y2)
Co-ordinates 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) = (x1, y1)
(-2, -2) = (x2, y2)
m1 : m2 = 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) = (x1, y1)
Let co-ordinates of other end be (x2, y2)
Now,
or, x = \(\frac{x_1 + x_2}{2}\)
or, 0 = \(\frac{-6 + x_2}{2}\)
or, 0 = - 6 + x2
\(\therefore\) x2 = 6
Again,
or, y = \(\frac{y_1 + y_2}{2}\)
or, 3 = \(\frac{8 + y_2}{2}\)
or, 6 = 8 + y2
or, 6 - 8 = y2
\(\therefore\) y2= -2

So, the co-ordinates of other end is (6, -2).

Solution:

Given,
(2, 2) = (x1, y1)
(-2,-2) = (x2, y2)
(x, y) = (\(\frac{6}{7}\), \(\frac{6}{7}\))
m1 : m2 = ?
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(m1+ m2) = 7(-2m1 + 2m2)
or, 6m1+ 6m2 = -14m1 + 14m2
or, 6m1+ 14m1=14m2 -6m2
or, 20m1 = 8m2
or, \(\frac{m_1}{m_2}\) = \(\frac{8}{20}\)
\(\therefore\) m1 : m2= 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) = (x1, y1)
Let the co-ordinates of the other end be (x2, y2)
or, x = \(\frac{x_1 + x_2}{2}\)
or, -4 = \(\frac{-5 + x_2}{2}\)
or, -8 = -5 + x2
or, -8 + 5 = x2
or, x2 = -3
Again,
or, y = \(\frac{y_1 + y_2}{2}\)
or, -5 = \(\frac{-3 + y_2}{2}\)
or, -10 = -3 + y2
or, y2 = -10 + 3
or, y2 = -7

\(\therefore\) The required point(x2, y2) = (-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 = -3m1 + 2m2
or, 3m1 = 2m2
or, \(\frac{m_1}{m_2}\) = \(\frac{2}{3}\)
\(\therefore\) m1 : m2 = 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) = (x1, y1)
(4, 8) = (x2, y2)
(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(m1 + m2) = 4m1 - 1m2
or, 2m1 + 2m2 = 4m1 - 1m2
or, 2m1- 4m1 = -1m2 - 2m2
or, -2m1 = -3m2
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)

0%
  • Find the co-ordinates of the mid point of the line segement joining the given points, (4, 6) and (2, 4).

    (3, 3)
    (3, 2)
    (3, 5)
    (5, 5)
  • Find the co-ordinates of the point (2, 3) and (7, 3) which divides the line segment joining the following points in the 3:2 ratio.

    4, 1
    3, 5
    2, 3
    5, 3
  • Find the ratio in which the y-axis 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 x-axis divides the line segment joining the points (3, 2) and (3, −9).

    2 : 3
    3 : 5
    2 : 9
    3 : 9
  • Find the co-ordinates of the mid point of the line segment joining the given points (5a, 7b) and (3a, −2b).

    (4, (frac{5b}{2}))
    (3, (frac{5b}{3}))
    (8, (frac{5b}{2}))
    (4, (frac{6b}{4}))
  • Find the co-ordinates of the point which divides the line segment joining the (2, −4) and (−3, 6) points in the 2:3 ratio.

    2, 1
    0, 0
    1, 2
    2, 3
  • Find the co-ordinates 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 co-ordinates of the mid point of the line segment joining the (8, 5) and (-12, -7) points.

    -2, 2, 
    1, 2
    -3, -1
    -2, -1
  • Find the co-ordinates 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 co-ordinates of the point which divides the line segment joining (2, 2) and (-2, -2) points in the 2:5 ratios.

    5, 6
    2 ,6
    ((frac{6}{7}), (frac{6}{7}))
    ((frac{5}{7}), (frac{6}{5}))
  • Find co-ordinates: (1, -3) and (4, 6), ratio 2:1

    3, 3
    1, 2
    1, 3
    2, 3
  • Find co-ordinates: (-3, -7) and (-5, -3)

    -4, 5
    4, 5
    -5, 4
    -4, -5
  • Find co-ordinates: (-6, 0) and (0, -8)

    -3, 4
    -4, -3
    3, 4
    -3, -4
  • Find co-ordinates: (-6, 8) and (6, -2)

    -3, 0
    0, 3
    1, -2
    3, 0
  • Find co-ordinates: (4, 6) and (2, 4)

    3, 5
    5, 3
    2, 3
    -3, -2
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Show that the points A (3,4) B(7,3) and C (4,8) are the vertices of an isosceles