A point moves so that the angle from the line

Step 1
Let the coordinate of the moving point be ${\displaystyle (\alpha ,\beta )}$ and origine ${\displaystyle (0,0)}$
Then slope of the line joining this two point ${\displaystyle (\alpha ,\beta )(0,0)}$ is
${\displaystyle m_1=\frac{\beta -0}{\alpha -0}=\frac{\beta }{\alpha }}$
and the slope of the line joining the point (3, -2) and (5, 7) is
${\displaystyle m_2=\frac{7-(-2)}{5-3}=\frac{9}{2}}$
Now given that the above two lins whose slope are ${\displaystyle m_1,m_2}$ makes an angle ${\displaystyle {45}^{\circ }}$
Then by angle between two lines formula
${\displaystyle {\tan 45}^{\circ }=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|}$
${\displaystyle \Rightarrow I=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|}$
When we remove moduls it will give + and -, (Taking + for actuale angle ${\displaystyle {45}^{\circ }}$)
When + Then ${\displaystyle 1=\frac{m_1-m_2}{1+m_1{m}_2}}$
${\displaystyle \Rightarrow 1+m_1{m}_2=m_1-m_2}$
${\displaystyle \Rightarrow 1+\frac{\beta }{\alpha }\times \frac{9}{2}=\frac{\beta }{\alpha }-\frac{9}{2}}$
${\displaystyle \Rightarrow 2\alpha +9\beta =2\beta -9\alpha }$
${\displaystyle \Rightarrow 11\alpha +7\beta =0}$
${\displaystyle \therefore }$ sins ${\displaystyle (\alpha ,\beta )}$ is arbitry
$\therefore$ locus of the point will be
${\displaystyle 11x+7y=0}$