If a particle moving along x-2y-3=0 gets refl | vedclass

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(B) The equation of the incident line is $x-2y-3=0$. The slope of this line is $m_1 = \frac{1}{2}$.Since the particle is reflected in a perpendicular

Vedclass · Accepted Answer

$2x+y-1=0$

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(B) The equation of the incident line is $x-2y-3=0$. The slope of this line is $m_1 = \frac{1}{2}$.Since the particle is reflected in a perpendicular direction,the slope of the reflected line $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$m_2 = -2$.The point of incidence is the intersection of $x-2y-3=0$ and $3x-2y-5=0$.Subtracting the first equation from the second: $(3x-2y-5) - (x-2y-3) = 0$ $\Rightarrow 2x-2=0$ $\Rightarrow x=1$.Substituting $x=1$ into $x-2y-3=0$,we get $1-2y-3=0$ $\Rightarrow -2y=2$ $\Rightarrow y=-1$.The point of intersection is $(1, -1)$.The equation of the reflected line passing through $(1, -1)$ with slope $m_2 = -2$ is:$y - (-1) = -2(x - 1)$$y + 1 = -2x + 2$$2x + y - 1 = 0$.

If a particle moving along $x-2y-3=0$ gets reflected in a perpendicular direction upon hitting the line $3x-2y-5=0$,then the line of the movement of the particle after reflection is

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