Question

A point moves so that the angle from the line joining it and the origin to the line (3, -2) and (5, 7) is 45º. Find the equation of the locus.

Answer 1

Solution:

Let the coordinates of moving point be P(h,k).

The equation of line joining P(h,k) and origin (0,0) is:

$y-k=\dfrac{k-0}{h-0}(x-h) \\\Rightarrow y-k=\dfrac{k}{h}(x-h) \\\Rightarrow y-k=\dfrac{kx}h-k \\\Rightarrow y=\dfrac{kx}h \ ...(i)$

Here, slope is $m_1=\dfrac kh$

Now, the equation of line joining (3,-2), (5,7) is:

$y-(-2)=\dfrac{7+2}{5-3}(x-3) \\\Rightarrow y+2=\dfrac{9}{2}(x-3) \\\Rightarrow y+2=\dfrac{9x}{2}-\dfrac{27}{2} \\\Rightarrow y=\dfrac{9x}{2}-\dfrac{31}{2}\ ...(ii)$

Here, slope is $m_2=\dfrac92$

Angle between the lines is $\theta$:

$\tan \theta=|\dfrac{m_1-m_2}{1+m_1m_2}| \\\Rightarrow \tan 45\degree=|\dfrac{\dfrac kh-\dfrac 92}{1+\dfrac kh\times \dfrac 92}|$

$\\\Rightarrow 1=|\dfrac{\dfrac {2k-9h}{2h}}{\dfrac {2h+9k}{2h}}| \\\Rightarrow \pm1=\dfrac {2k-9h}{2h+9k}$

$\\\Rightarrow 2h+9k=2k-9h;\ \ 2h+9k=9h-2k \\\Rightarrow 11h+7k=0; \ \ 7h-11k=0 \\\Rightarrow 11x+7y=0; \ \ 7x-11y=0$