The equation of a straight line passing through the point $(1,2)$ and inclined at $45^{\circ}$ to the line…
The equation of a straight line passing through the point $(1,2)$ and inclined at $45^{\circ}$ to the line $y=2 x+1$ is
- $5 x+y=7$
- $3 x+y=5$
- $x+y=3$
- $x-y+1=0$
Solution
Let required equation of line is
$y=m x+c$
Since, angle between $y=m x+c$ and
$y=2 x+1$ is $\tan 45^{\circ}$
$\begin{array}{ll}\because \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \Rightarrow & \tan 45^{\circ}=\left|\frac{m-2}{1+2 m}\right| \\ \Rightarrow & 1= \pm \frac{m-2}{1+2 m} \\ \Rightarrow & 1+2 m=m-2 \\ \text { or } & 1+2 m=-m+2 \\ \Rightarrow & m=-3 \text { or } 3 m=1 \\ \Rightarrow & m=-3 \text { or } \frac{1}{3}\end{array}$
On putting $m=-3$ in Eq. (i), we get
$y=-3 x+c$
Since, it passes through $(1,2)$.
$\begin{aligned} & \therefore \quad 2=-3+c \\ & \Rightarrow \quad c=5 \\ & \therefore \quad y=-3 x+5 \\ & \Rightarrow \quad 3 x+y=5 \\ & \end{aligned}$
Again, putting $m=\frac{1}{3}$ in Eq. (i), we get
$y=-\frac{x}{3}+c$
Since, it passes through $(1,2)$.
$\therefore \quad 2=-\frac{1}{3}+c \Rightarrow c=\frac{7}{6}$
$\therefore$ Equation of line is $y=-\frac{x}{3}+\frac{7}{6}$
$\Rightarrow \quad 6 y+2 x=7$
Hence, option (b) is correct.
Asked in: AP EAMCET 2012
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