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The slope of lines which makes an angle $45^{\circ}$ with the line $2 x-y=-7$
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Verified Answer
The correct answer is:
$\frac{1}{3},-3$
We have, $\theta=\frac{\pi}{4}$
Given, line is $2 x-y=-7$
$$
y=2 x+7
$$
Here, slope $m_1=2$
The required line makes an angle $45^{\circ}$ with this line
So, $\tan 45^{\circ}=\left|\frac{2-m_2}{1+2 m_2}\right|$
$\Rightarrow \quad 1=\left|\frac{2-m_2}{1+2 m_2}\right|$
$\Rightarrow 1+2 m_2=2-m_2$ or $1+2 m_2=-\left(2-m_2\right)$
$2 m_2+m_2=2-1$ or $1+2 m_2=-2+m_2$
$\Rightarrow \quad 3 m_2=1 \quad \Rightarrow 2 m_2-m_2=-2-1$
$\therefore m_2=\frac{1}{3}, m_2=-3$
Given, line is $2 x-y=-7$
$$
y=2 x+7
$$
Here, slope $m_1=2$
The required line makes an angle $45^{\circ}$ with this line
So, $\tan 45^{\circ}=\left|\frac{2-m_2}{1+2 m_2}\right|$
$\Rightarrow \quad 1=\left|\frac{2-m_2}{1+2 m_2}\right|$
$\Rightarrow 1+2 m_2=2-m_2$ or $1+2 m_2=-\left(2-m_2\right)$
$2 m_2+m_2=2-1$ or $1+2 m_2=-2+m_2$
$\Rightarrow \quad 3 m_2=1 \quad \Rightarrow 2 m_2-m_2=-2-1$
$\therefore m_2=\frac{1}{3}, m_2=-3$
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