$$\begin{gathered} ( \mathrm{i} ) \mathrm{Slope} \mathrm{of} \mathrm{PQ} = \frac{3 - \mathrm{k}}{1 - 3 \mathrm{k} - 6} \\ \Rightarrow \frac{1}{2} = \frac{3 - \mathrm{k}}{- 3 \mathrm{k} - 5} \\ \Rightarrow - 3 \mathrm{k} - 5 = 2 ( 3 - \mathrm{k} ) \\ \Rightarrow - 3 \mathrm{k} - 5 = 6 - 2 \mathrm{k} \\ \Rightarrow \mathrm{k} = - 11 \end{gathered}$$

( ii ) Substituting  k  in  p  and  Q,  we get

P ( 6, K ) = P ( 6, - 11 )   and 

Q ( 1 - 3 k, 3 )  =  Q ( 1 - 3 ( - 11 ), 3 )  =  Q ( 1 +  33, 3 )  =  Q ( 34, 3 )

$$\begin{gathered} \therefore \mathrm{Midpoint} \mathrm{of} \mathrm{PQ} = \left( \frac{6 + 34}{2}, \frac{- 11 + 3}{2} \right) \\ = \left( \frac{40}{2}, \frac{- 8}{2} \right) \\ = ( 20, - 4 ) \end{gathered}$$

(i) Given points are A(- 4, 2) and B(3, 6)

Let P(x, 3) divides the line joining A(- 4,2) and B(3,6) in the ratio k :1.

Thus, we have

$\frac{3\mathrm{k} - 4}{\mathrm{k} + 1} = \mathrm{x}$      ...(i)

And $\frac{6\mathrm{k} + 2}{\mathrm{k} + 1} = 3$

$\Rightarrow$6k + 2 = 3(k + 1)

$\Rightarrow$ 6k + 2 = 3k + 3

$\Rightarrow$ 3k = 1

$\Rightarrow \mathrm{k} = \frac{1}{3}$

By substituting the value of k in equation(i), we have

$\frac{{\displaystyle 3 \times \frac{1}{3} - 4}}{{\displaystyle \frac{1}{3} + 1}} = \mathrm{x}$

$\Rightarrow \frac{- 3}{{\displaystyle \frac{4}{3}}} = \mathrm{x}$

$\Rightarrow \frac{- 9}{4} = \mathrm{x}$

Thus, coordinates of P are $(\frac{- 9}{4}, 3)$

(ii) Now, the distance AP = $\sqrt{(\frac{- 9}{4} + 4{)}^2 + (3 - 2{)}^2}$

                                   = $\sqrt{(\frac{7}{4}{)}^2 + {\left(1\right)}^2}$

                                  = $\sqrt{\frac{49}{16} + 1}$

                                  = $\sqrt{\frac{49 + 16}{16}}$

                                  = $\sqrt{\frac{65}{16}}$

$\Rightarrow$ AP = $\frac{\sqrt{65}}{4}$ units

Given,

The coordinates of the vertices of $∆$ABC are A(1, 3), B(4, b) and C(a, 1).

The coordinates of the centroid are G(4, 3).

It is known that A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are vertices of a triangle, then

Thus, the coordinates of the centroid of $∆$ABC are ($\frac{{\mathrm{x}}_1 + {\mathrm{x}}_{2 }+ {\mathrm{x}}_3}{3}, \frac{{\displaystyle {\mathrm{y}}_1 +{\mathrm{y}}_{2 }+ {\mathrm{y}}_3}}{{\displaystyle 3}}$)

$\Rightarrow$ the coordinates of the centroid of $∆$ABC are ($\frac{1 + 4 + \mathrm{a}}{3}, \frac{3 + \mathrm{b} + 1}{3}$) = ($\frac{5 + \mathrm{a}}{3}, \frac{4 + \mathrm{b}}{3}$)

But, as given the coordinates of the centroid are G(4, 3).

Thus, we have

$\frac{5 + \mathrm{a}}{3} = 4 \mathrm{and} \frac{4 +\hspace{0.17em}\mathrm{b}}{3} = 3$

$\Rightarrow 5 + \mathrm{a} =12 \mathrm{and} 4 + \mathrm{b} = 9$

$\Rightarrow \mathrm{a} = 7 \mathrm{and} \mathrm{b} = 5$

Thus, the coordinates of B and C are (4, 5) and (7, 1) respectively.

Using distance formula, we have

BC = $\sqrt{(7 - 4{)}^2 + (1 - 5{)}^2}$

     = $\sqrt{9 +\hspace{0.17em}16}$

     = $\sqrt{25}$

     = 5 units.

Thus, the length of BC is 5 units.

Given,  P ( 1, - 2 ),  A ( 3, - 6 )  and  B ( x, y )

 AP : PB  = 2 : 3

$$\begin{gathered} \mathrm{Hence}, \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{P} = \left( \frac{( 2 \times \mathrm{x} ) + ( 3 \mathrm{x} 3 )}{2 + 3}, \frac{ ( 2 \times \mathrm{y} ) + ( 3 \times \left( - 6 \right) )}{2 + 3} \right) \\ = \left( \frac{2 \mathrm{x} + 9}{5}, \frac{2 \mathrm{y} - 18}{5} \right) \end{gathered}$$

But,  the co-ordinates of  P  are  ( 1, - 2 ).

$$\begin{gathered} \therefore \frac{2 \mathrm{x} + 9}{5} = 1 \mathrm{and} \frac{2 \mathrm{y} - 18}{5} = - 2 \\ \Rightarrow 2 \mathrm{x} + 9 = 5 \mathrm{and} 2 \mathrm{y} - 18 = - 10 \\ \Rightarrow 2 \mathrm{x} = 5 - 9 \mathrm{and} 2 \mathrm{y} = - 10 + 18 \\ \Rightarrow \mathrm{x} = \frac{- 4}{2} \mathrm{and} \mathrm{y} = \frac{8}{2} \\ \Rightarrow \mathrm{x} = - 2 \mathrm{and} \mathrm{y} = 4 \end{gathered}$$

Hence,  the  co-ordinates of  B  are  ( - 2, 4 ).

a) Let the coordinates of A and B be (x, 0) and (0, y) respectively.

Given P divides AB is the ratio 2:3,

Using bisection formula, we have

$- 3 = \frac{2 \times 0 + 3 \times \mathrm{x}}{2 + 3} \mathrm{and} 4 = \frac{2 \times \mathrm{y} + 3 \times 0}{2 + 3}$

$- 3 = \frac{ 3\mathrm{x}}{5} \mathrm{and} 4 = \frac{2\mathrm{y}}{5}$

$- 15 = 3\mathrm{x} \mathrm{and} 20 = 2\mathrm{y}$

$x =- 5 \mathrm{and} y = 10$

Thus, the coordinates of A and B are (–5, 0) and (0, 10) respectively.