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If a particle moving along $x-2 y-3=0$ gets reflected in a perpendicular direction upon hitting the line $3 x-2 y-5=0$, then the line of the movement of the particle after reflection is
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The correct answer is:
$2 x+y-1=0$
Given line in $x-2 y-3=0$ which reflected by the line $3 \mathrm{x}-2 \mathrm{y}-5=0$.
Let the slope of reflected ray in $\mathrm{m}^1$
$$
2 \mathrm{y}=\mathrm{x}-3
$$
$$
y=\frac{1}{2} x-\frac{3}{2}
$$
$$
\mathrm{m}_1=\frac{1}{2}
$$
As inclined ray and reflected ray are normal to each other then, $\mathrm{mm}^1=-1$.
So, $\mathrm{m}^1=-2$.
Intersection of two lines $x-2 y-3=0$ and $3 x-2 y-5$ $=0$ is $(1,-1)$.
Required equation of line $=(y+1)=-2(x-1)$
$$
\Rightarrow 2 \mathrm{x}+\mathrm{y}-1=0
$$
Let the slope of reflected ray in $\mathrm{m}^1$
$$
2 \mathrm{y}=\mathrm{x}-3
$$
$$
y=\frac{1}{2} x-\frac{3}{2}
$$
$$
\mathrm{m}_1=\frac{1}{2}
$$
As inclined ray and reflected ray are normal to each other then, $\mathrm{mm}^1=-1$.
So, $\mathrm{m}^1=-2$.
Intersection of two lines $x-2 y-3=0$ and $3 x-2 y-5$ $=0$ is $(1,-1)$.
Required equation of line $=(y+1)=-2(x-1)$
$$
\Rightarrow 2 \mathrm{x}+\mathrm{y}-1=0
$$
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