Question

Equations of the two straight lines passing through the point \[\left( 3,2 \right)\] and making an angle of \[45{}^\circ \] with the line \[x-2y=3\], are
A. \[3x+y+7=0\] and \[x+3y+9=0\]
B. \[3x-y-7=0\] and \[x+3y-9=0\]
C. \[x+3y-7=0\] and \[x+3y-9=0\]
D. None of these

Answer 1

Hint: In this question, we are to find the equations of the two straight lines passing through the given point and making an angle of \[45{}^\circ \] with the given line. Here the acute angle formula is applied in order to calculate the slopes if the required lines. By substituting the slope and the point in the point-slope formula, we get the required straight lines.

Formula Used:The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
We can write the above equation in the point-slope form as
$y-{{y}_{1}}=m(x-{{x}_{1}})$
The slope of a line with the equation $ax+by+c=0$ is $m=\frac{-a}{b}$
The acute angle $\theta $ between the lines with slopes \[{{m}_{1}},{{m}_{2}}\] is $\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$



Complete step by step solution:It is given that,
Two straight lines that pass through the point $({{x}_{1}},{{y}_{1}})=(3,2)$ and makes an angle of \[45{}^\circ \] with the given line \[x-2y=3\].
The slope of the given line is
 $\begin{align}
  & {{m}_{2}}=\frac{-a}{b} \\
 & \text{ }=\frac{-1}{-2} \\
 & \text{ }=\frac{1}{2} \\
\end{align}$
Then, the slope of the required line is calculated by using the acute angle formula we have
$\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$
Here we have \[\theta =45{}^\circ \] and \[{{m}_{2}}=\frac{1}{2}\].
On substituting,
$\begin{align}
  & \tan 45{}^\circ =\left| \frac{m-\frac{1}{2}}{1+m\frac{1}{2}} \right| \\
 & \text{ }\Rightarrow 1=\left| \frac{2m-1}{2+m} \right| \\
 & \text{ }\Rightarrow 1=\frac{\left| 2m-1 \right|}{\left| 2+m \right|} \\
 & \Rightarrow \left| 2+m \right|=\left| 2m-1 \right| \\
\end{align}$
Squaring on both sides, we get
$\begin{align}
  & {{(2+m)}^{2}}={{(2m-1)}^{2}} \\
 & \Rightarrow 4+{{m}^{2}}+4m=4{{m}^{2}}-4m+1 \\
 & \Rightarrow 3{{m}^{2}}-8m-3=0 \\
\end{align}$
On simplifying, we get
$\begin{align}
  & 3{{m}^{2}}-9m+m-3=0 \\
 & \Rightarrow 3m(m-3)+1(m-3)=0 \\
 & \Rightarrow (m-3)(3m+1)=0 \\
\end{align}$
$\therefore m=3;\frac{-1}{3}$
If the value of the slope $m=3$, then the required equation of the line passing through $({{x}_{1}},{{y}_{1}})=(3,2)$ is
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & \Rightarrow y-2=3(x-3) \\
 & \Rightarrow y-2=3x-9 \\
 & \therefore 3x-y-7=0\text{ }...(1) \\
\end{align}$
If the value of the slope $m=\frac{-1}{3}$, then the required equation of the line passing through $({{x}_{1}},{{y}_{1}})=(3,2)$ is
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & \Rightarrow y-2=\frac{-1}{3}(x-3) \\
 & \Rightarrow 3y-6=-x+3 \\
 & \therefore x+3y-9=0\text{ }...(2) \\
\end{align}$
Therefore, equations (1) and (2) are the required straight lines.



Option ‘B’ is correct

Note: Here we may go wrong with the calculation of the slope. We use the acute angle formula since the required straight lines make an acute angle with the given line. Sometimes we may get a single slope and a single equation. In this way, the required equations are calculated.