Question

The equation of the line passing through (1,1) and makes an angle 45° with the line 2x-y+7=0

Answer 1

The angle between two lines $\theta$with slopes $m_1$ and $m_2$ is such that

$\tan \theta=\pm \frac{m_1-m_2}{1+m_1m_2}$

Here $\theta=45^o$ and let the slope of the line $2x-y+7=0$ be $m_2=2$, then

$\tan 45^o=\pm \frac{m_1-2}{1+2m_1} \\ 1=\pm \frac{m_1-2}{1+2m_1} \\ 1+2m_1= \pm (m_1-2)\\ 1+2m_1=m_1-2 \; or\; 1+2m_1=2-m_1\\ m_1=-3,\frac{1}{3}$

The equation of line passing through (1,1) with slope $m_1$ is

$\frac{y-1}{x-1} =m_1$, or we can write,

$\frac{y-1}{x-1} =-3 \;or \; \frac{y-1}{x-1} =\frac{1}{3} \\ 3x+y-4=0 \; or \; x-3y+2=0$

Thu the possible equations of the lines satisfying the given conditions are

$3x+y-4=0 \; or \; x-3y+2=0$.

Answer 2

so answer is 3x + y - 4=0 or x -3y + 2 = 0