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)