Question

A line PQ makes intercepts of length 2 units between the lines $y + 2x = 3{\text{ and }}y + 2x = 5.$ If the coordinates of P are (2,3), coordinates of Q can be
A) $\left( {6,0} \right)$
B) $\left( {2,3} \right)$
C) $\left( {0,\dfrac{9}{2}} \right)$
D) $\left( {3,2} \right)$

Answer 1

Hint: Here, we solve this question by using the concept of straight lines. In this question the lines given are parallel lines. Parallel lines have the same slope and will never intersect.
Two lines ${{\text{L}}_{\text{1}}}{\text{: y = }}{{\text{m}}_{\text{1}}}{\text{x + }}{{\text{c}}_{\text{1}}}{\text{ and }}{{\text{L}}_{\text{2}}}{\text{: y = }}{{\text{m}}_{\text{2}}}{\text{x + }}{{\text{c}}_{\text{2}}}$ are said to be parallel if ${m_1} = {m_2}$
Distance between two parallel lines is $D = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
For line passing through a given point $\left( {{x_1},{y_1}} \right)$and having slope m,
$y - {y_1} = m\left( {x - {x_1}} \right)$
When the line passes through origin, then such a line is represented by y = mx in point-slope form.

Complete step-by-step answer:
As the lines ${\text{y + 2x = 3 and y + 2x = 5}}$ are having the same slope, hence they are parallel to each other.
Since PQ makes an intercept of 2 units between the parallel lines.
Distance between the parallel lines is D.
$\begin{gathered}
  D = \left| {\dfrac{{{c_2} - {c_1}}}{{\sqrt {{A^2} + {B^2}} }}} \right| \\
  D = \left| {\dfrac{{5 - 3}}{{\sqrt {4 + 1} }}} \right| \\
  D = \dfrac{2}{{\sqrt 5 }} \\
\end{gathered} $
As tan θ is called slope of the straight line i.e. \[m = {\text{ }}tan\theta \]
${\text{tan\theta = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ and sin\theta = }}\dfrac{{\text{1}}}{{\sqrt {\text{5}} }}$
θ is the angle between the lines and the line passing through the PQ.
Now,
-2 is the slope of given parallel lines
As angle between straight lines, $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$
Let the slope of line PQ be m
$\begin{gathered}
  \dfrac{1}{2} = \dfrac{{\left| { - 2 - m} \right|}}{{\left| {1 - 2m} \right|}} \\
   \Rightarrow \dfrac{{ - 2 - m}}{{1 - 2m}} = \pm \dfrac{1}{2} \\
   \Rightarrow \dfrac{{ - 2 - m}}{{1 - 2m}} = \dfrac{1}{2}{\text{ and }}\dfrac{{ - 2 - m}}{{1 - 2m}} = - \dfrac{1}{2} \\
\end{gathered} $
$ \Rightarrow \dfrac{{ - 2 - m}}{{1 - 2m}} = \dfrac{1}{2}$, on solving this we will not get the value of m
So, on solving $\dfrac{{ - 2 - m}}{{1 - 2m}} = - \dfrac{1}{2}$
$\begin{gathered}
  2\left( { - 2 - m} \right) = 1 - 2m \\
   - 4 - 2m = - \left( {1 - 2m} \right) \\
   - 4 - 2m = - 1 + 2m \\
  4m = - 3 \\
  m = - \dfrac{3}{4} \\
\end{gathered} $
Therefore, the equation of the line PQ
$\begin{gathered}
   - \dfrac{3}{4} = \dfrac{{y - 3}}{{x - 2}} \\
   - 3\left( {x - 2} \right) = 4\left( {y - 3} \right) \\
   - 3x + 6 = 4y - 12 \\
  3x + 4y = 18 \\
\end{gathered} $
Verifying through options, we get the coordinates of Q as (6, 0) and $\left( {0,\dfrac{9}{2}} \right)$

Note: To find the relation between two lines:
Let L1and L2 be the two lines as
$\begin{gathered}
  {L_1}:{a_1}x + {b_1}y + {c_1} = 0 \\
  {L_2}:{a_2}x + {b_2}y + {c_2} = 0 \\
\end{gathered} $
For Parallel lines : Two lines are said to be parallel if satisfy the below condition,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
For Intersection of lines : Two lines intersect a point if
$\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$
For Coincident Lines : Two lines intersect a point if
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$