Question

If the three lines $x - 3y = p,ax + 2y = q$ and $ax + y = r$ form a right-angled triangle then:
This question has multiple correct options
A.$9{a^2} + 12a + 4 = 0$
B.$9{a^2} + 6a + 1 = 0$
C.${a^2} - 6a - 18 = 0$

Answer 1

Hint: Here we will use the line equation, slope formula and the relation between the slopes of the two perpendicular lines. Follow the concept that out of the given three lines any line can form the right angle with the other line.

Complete step-by-step answer:
Take the given three lines.
$
  x - 3y = p\;{\text{ }}....{\text{ (i)}} \\
  ax + 2y = q\;{\text{ }}....{\text{ (ii)}} \\
 $
And
$ax + y = r\,{\text{ }}.....{\text{ (iii)}}$
Therefore by using standard line equation –
$y = mx + c$ and slope $m = \dfrac{{ - a}}{b}$
Given that the three lines form right angle.
Therefore any of the two lines are perpendicular to each other.
Either line (i) is perpendicular to line (ii) or line (ii) is perpendicular to line (iii)
Where, slope of the two perpendicular lines is
$
  {m_1}{m_2} = ( - 1) \\
   \Rightarrow {m_1} = \dfrac{{( - 1)}}{{{m_2}}} \\
 $
First take perpendicular lines for lines (i) and (ii)
$ \Rightarrow \dfrac{{ - 1}}{{ - 3}} = \dfrac{{ - a}}{2}$
Negative sign on the numerator and the denominator cancels each other on the left hand side of the equation-
$ \Rightarrow \dfrac{1}{3} = \dfrac{{ - a}}{2}$
Simplify the above equation by cross-multiplication-
$ \Rightarrow 2 = - 3a$
When the term changes its side from right hand side to left side then the sign of the term also changes from negative to positive and vice-versa.
$ \Rightarrow 3a + 2 = 0$
Take whole square on both the sides of the equation-
$ \Rightarrow {\left( {3a + 2} \right)^2} = 0$
Simplify the above equation –
$ \Rightarrow 9{a^2} + 12a + 4 = 0\,{\text{ }}.....{\text{ (iv)}}$
Similarly take perpendicular for lines (ii) and (iii)-
$ \Rightarrow \dfrac{1}{3} = - a$
Simplify the above equation by cross-multiplication-
$ \Rightarrow 1 = - 3a$
When the term changes its side from right hand side to left side then the sign of the term also changes from negative to positive and vice-versa.
$ \Rightarrow 3a + 1 = 0$
Take whole square on both the sides of the equation-
$ \Rightarrow {\left( {3a + 1} \right)^2} = 0$
Simplify the above equation –
$ \Rightarrow 9{a^2} + 6a + 1 = 0\,{\text{ }}.....{\text{ (v)}}$
Hence, from the equations (iv) and (v), we can say that from the given multiple choices – the option A and B are the correct answer.

Note: Be careful while solving the slope of the perpendicular lines. Be careful while dealing with negative and positive terms and moving terms from one side to another.