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 (\(\frac{6}{7}\), \(\frac{6}{7}\))

Solution:

(0, 3) = (x, y)

(-6, 8) = (x_{1}, y_{1})

Let co-ordinates 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 co-ordinates 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 co-ordinates 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)