Question

Through the point $(3,4)$ are drawn two straight lines each inclined at ${45}^o$ to the straight line$x-y=2$. Find their equations and find also the area included by the three lines.

Answer 1

As $tan\theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

Given $\theta ={45}^0$ and slope $m_2=1$ (From given equation in the question)

Then, $tan{45}^0=1=\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

$\Rightarrow$ $\frac{1-m_1}{1+m_1}=1$ or $\frac{1-m_1}{1+m_1}=-1$

$\Rightarrow$ $m_1=0$ or $m_1=\infty$

Let $y=mx+c$ be a line then

For $m_1=0$ : $y=c$ and given this line passes through $(3,4)$ then equation will be $y=4$.........$\left(1\right)$

For $m_1=\infty$ : $x=c$ and given this line passes through $(3,4)$ then equation will be $x=3$...........$\left(2\right)$

Then area of these three lines $\left(1\right)$, $\left(2\right)$ and $x-y=2$ is

$A=\frac{1}{2}\times 3\times 3=\frac{9}{2}sq.units$...(we get co-ordinate points by solving $\left(1\right),\left(2\right),\left(3\right)$ and by using are formulae we get this)