# A point moves so that the angle from the line

A point moves so that the angle from the line joining it and the origin to the line (3, - 2) and (5, 7) is ${45}^{\circ }$ Find the equation of the locus.
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Jacob Homer

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

###### Not exactly what you’re looking for?
Donald Cheek
Step 1
Then, according to the formula for the angle between two lines
${\mathrm{tan}45}^{\circ }=|\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}|$
$I=|\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}|$
If we remove modules, there will be + and -
Then $1=\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}$
$1+{m}_{1}{m}_{2}={m}_{1}-{m}_{2}$
$1+\frac{\beta }{\alpha }×\frac{9}{2}=\frac{\beta }{\alpha }-\frac{9}{2}$
$2\alpha +9\beta -2\beta -9\alpha$
$11\alpha +7\beta =0$
The locus of the point will be
$11x+7y=0$