If the straight lines 3x – 5y = 7 and 4x + ay + 9 = 0 are perpendicular to one another, find the value of a.

3 x - 5 y = 7

4 x + a y + 9 = 0

$$\begin{gathered} \Rightarrow \mathrm{a} \mathrm{y} = - 4 \mathrm{x} - 9 \\ \Rightarrow \mathrm{y} = \frac{- 4}{\mathrm{a}} \mathrm{x} - \frac{9}{\mathrm{a}} \\ \Rightarrow \mathrm{Its} \mathrm{slope} = \frac{- 4}{\mathrm{a}} \end{gathered}$$

Since lines are perpendicular to each other,

$$\begin{gathered} \frac{3}{5} \times \frac{{\displaystyle - 4}}{{\displaystyle \mathrm{a}}} = - 1 \Rightarrow \frac{{\displaystyle 3}}{{\displaystyle 5}} \times \frac{{\displaystyle 4}}{{\displaystyle \mathrm{a}}} = 1 \\ \Rightarrow \frac{{\displaystyle 4}}{{\displaystyle \mathrm{a}}} = \frac{5}{3} \\ \Rightarrow \mathrm{a} = \frac{4 \times 3}{5} \\ \Rightarrow \mathrm{a} = \frac{12}{5} \end{gathered}$$

let the co-ordinates of M be ( x, y ).

Thus we have

$$\begin{gathered} \mathrm{x} = \frac{{\mathrm{m}}_1 {\mathrm{x}}_2 + {\mathrm{m}}_2 {\mathrm{x}}_1}{{\mathrm{m}}_1 + {\mathrm{m}}_2} \\ = \frac{1 \times \left( - 1 \right) + 2 \times 2}{1 + 2} \\ = \frac{- 1 + 4}{3} = \frac{3}{3} \\ \mathrm{x} = 1 \end{gathered}$$

$$\begin{gathered} \mathrm{y} = = \frac{{\mathrm{m}}_1 {\mathrm{y}}_2 + {\mathrm{m}}_2 {\mathrm{y}}_1}{{\mathrm{m}}_1 + {\mathrm{m}}_2} \\ = \frac{1 \times \left( 2 \right) + 2 \times 5}{1 + 2} \\ = \frac{2 + 10}{3} = \frac{12}{3} \\ \mathrm{y} = 4 \\ \Rightarrow \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{M} \mathrm{are} \left( 1, 4 \right). \end{gathered}$$

$$\begin{gathered} \mathrm{Slope} \mathrm{of} \mathrm{line} \mathrm{passing} \mathrm{through} \mathrm{C} \mathrm{and} \mathrm{M} = \mathrm{m} = \frac{4 - 8}{1 - 5} = \frac{- 4}{- 4} = 1 \\ \therefore \mathrm{Required} \mathrm{equation} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{y} - 8 = 1 ( \mathrm{x} - 5 ) \\ \Rightarrow \mathrm{y} - 8 = \mathrm{x} - 5 \\ \Rightarrow \mathrm{y} = \mathrm{x} + 3. \end{gathered}$$

The image of point  ( x, y )  0n  y-axis has the co-ordinates  ( - x, y ).

Thus,  we have

Co-ordinates of  B' = ( - 2, 3 )

Co-ordinates of  C' = ( - 1, 1 )

Co-ordinates of  D' = ( - 2, 0 )

Since,  Y-axis is the line of symmetry of the figure formed, the equation of the line of symmetry is  x = 0.

Given vertices:   A ( - 1, 3 ),  B ( 4, 2 )  and  C ( 3, - 2 )

( i ) Co-ordinates of the centroid  G  of $∆ \mathrm{ABC}$ are given by

$$\begin{gathered} \mathrm{G} = \left( \frac{- 1 + 4 + 3}{3}, \frac{3 + 2 - 2}{3} \right) \\ = \left( \frac{6}{3}, \frac{3}{3} \right) = ( 2, 1 ) \end{gathered}$$

( ii ) Since the line through  G  is parallel to  AC,  the slope of the lines are the same.

$$\begin{gathered} \Rightarrow \mathrm{m} \frac{{\mathrm{y}}_2 - {\mathrm{y}}_1}{{\mathrm{x}}_2 - {\mathrm{x}}_1} = \frac{- 2 - 3}{3 - \left( - 1 \right)} = \frac{- 5}{4} \\ \mathrm{So}, \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{line} \mathrm{passing} \mathrm{through} \mathrm{G} ( 2, 1 ) \mathrm{and} \mathrm{with} \mathrm{slope} \frac{- 5}{4} \mathrm{is} \mathrm{given} \mathrm{by}, \\ \mathrm{y} - {\mathrm{y}}_1 = \mathrm{m} ( \mathrm{x} - {\mathrm{x}}_1 ) \\ \Rightarrow \mathrm{y} - 1 = \frac{- 5}{4} \left( \mathrm{x} - 2 \right) \\ \Rightarrow 4 \mathrm{y} - 4 = - 5 \mathrm{x} + 10 \\ \Rightarrow 5 \mathrm{x} + 4 \mathrm{y} = 14 \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{equation}. \end{gathered}$$

(i) Since,  A  lies on the  x-axis, let the coordinates of  A  be  (x, 0).

Since  B  lies on the  y - axis, let the coordinates of  B  be  (0, y).

Let  m = 1   and    n = 2.

Using section formula, 

$$\begin{gathered} \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{P} = \left( \frac{1 ( 0 ) + 2 ( \mathrm{x} )}{1 + 2}, \frac{1 \mathrm{y} + 2 ( 0 )}{1 + 2} \right) \\ \Rightarrow ( 4, - 1 ) = \left( \frac{2 \mathrm{x}}{3}, \frac{\mathrm{y}}{3} \right) \\ \Rightarrow \frac{2 \mathrm{x}}{3} = 4 \mathrm{and} \frac{\mathrm{y}}{3} = - 1 \\ \Rightarrow 2 \mathrm{x} = 3 \mathrm{x} 4 \mathrm{and} \mathrm{y} = - 1 \mathrm{x} 3 \\ \Rightarrow \mathrm{x} = \frac{12}{2} \mathrm{and} \mathrm{y} = - 3 \\ \Rightarrow \mathrm{x} = 6 \mathrm{and} \mathrm{y} = - 3 \end{gathered}$$

So, the co-ordinates of  A  are ( 6, 0 )  and that of  B  are  (0, - 3 ).

$$\begin{gathered} ( \mathrm{ii} ) \mathrm{Slope} \mathrm{of} \mathrm{AB} = \frac{- 3 - 0}{0 - 6} = \frac{- 3}{- 6} = \frac{1}{2} \\ \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{line} \mathrm{perpendicular} \mathrm{to} \mathrm{AB} = \mathrm{m} =- 2 \\ \mathrm{P} ( 4, - 1 ) \\ \Rightarrow \mathrm{Required} \mathrm{equation} \mathrm{is} \\ \mathrm{y} - {\mathrm{y}}_1 = \mathrm{m} ( \mathrm{x} - {\mathrm{x}}_1 ) \\ \Rightarrow \mathrm{y} - ( - 1 ) = - 2 ( \mathrm{x} - 4 ) \\ \Rightarrow \mathrm{y} + 1 = - 2 \mathrm{x} + 8 \\ \Rightarrow 2 \mathrm{x} + \mathrm{y} = 7. \end{gathered}$$