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 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, 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.
Solution:
Here,
(4, 6) = (x_{1}, y_{1})
(2, 4) = (x_{2}, y_{2})
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) = (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 y-axis 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 y-axis 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})
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) = (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) = (\(\frac{- 6+ 6}{5}\), \(\frac{12-12}{5}\))
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})
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) = (x_{1}, y_{1})
(-5, -3) = (x_{2}, y_{2})
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) = (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 (\(\frac{6}{7}\), \(\frac{6}{7}\))
Find the co-ordinates of the mid point of the line segement joining the given points, (4, 6) and (2, 4).
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.
Find the ratio in which the y-axis divides the line segment joining the points (−4, 1) and (10, 1)
Find the ratio in which x-axis divides the line segment joining the points (3, 2) and (3, −9).
Find the co-ordinates of the mid point of the line segment joining the given points (5a, 7b) and (3a, −2b).
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.
Find the co-ordinates of the point which divides the line segment joining (-10, 12) and (-3, -9) points in the 4:3 ratio.
Find the co-ordinates of the mid point of the line segment joining the (8, 5) and (-12, -7) points.
Find the co-ordinates of the mid point of the line segment joining (-3, -7) and (-5, -3) points.
Find the co-ordinates of the point which divides the line segment joining (2, 2) and (-2, -2) points in the 2:5 ratios.
Find co-ordinates: (1, -3) and (4, 6), ratio 2:1
Find co-ordinates: (-3, -7) and (-5, -3)
Find co-ordinates: (-6, 0) and (0, -8)
Find co-ordinates: (-6, 8) and (6, -2)
Find co-ordinates: (4, 6) and (2, 4)
You must login to reply
pankaj
Show that the points A (3,4) B(7,3) and C (4,8) are the vertices of an isosceles
Jan 01, 2017
1 Replies
Successfully Posted ...
Please Wait...