JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 17)
A ray of light is incident along a line which meets another line, 7x − y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is :
41x − 38y + 38 = 0
41x + 25y − 25 = 0
41x + 38y − 38 = 0
41x − 25y + 25 = 0
Explanation
Let slope of incident ray be m.
$$ \therefore $$ angle of incidence = angle of reflection
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$$ \therefore $$ $$\left| {{{m - 7} \over {1 + 7m}}} \right| = \left| {{{ - 2 - 7} \over {1 - 14}}} \right| = {9 \over {13}}$$
$$ \Rightarrow $$ $${{m - 7} \over {1 + 7m}} = {9 \over {13}}$$
or $${{m - 7} \over {1 + 7m}} = - {9 \over {13}}$$
$$ \Rightarrow $$ 13m $$-$$ 91 $$=$$ 9 + 63m
or 13m $$-$$ 91 $$=$$ $$-$$ 9 $$-$$ 63m
$$ \Rightarrow $$ 50m $$=$$ $$-$$ 100 or 76m $$=$$ 82
$$ \Rightarrow $$ m $$=$$ $$ - {1 \over 2}$$
or m $$=$$ $${{41} \over {38}}$$
$$ \Rightarrow $$ y $$-$$ 1 $$=$$ $$-$$ $${1 \over 2}$$ (x $$-$$ 0)
or y $$-$$ 1 $$=$$ $${{41} \over {38}}$$ (x $$-$$ 0)
i.e x + 2y $$-$$ 2 $$=$$ 0
or 38y $$-$$ 38 $$-$$ 41x $$=$$ 0
$$ \Rightarrow $$ 41x $$-$$ 38y + 38 $$=$$ 0
$$ \therefore $$ angle of incidence = angle of reflection
$$ \therefore $$ $$\left| {{{m - 7} \over {1 + 7m}}} \right| = \left| {{{ - 2 - 7} \over {1 - 14}}} \right| = {9 \over {13}}$$
$$ \Rightarrow $$ $${{m - 7} \over {1 + 7m}} = {9 \over {13}}$$
or $${{m - 7} \over {1 + 7m}} = - {9 \over {13}}$$
$$ \Rightarrow $$ 13m $$-$$ 91 $$=$$ 9 + 63m
or 13m $$-$$ 91 $$=$$ $$-$$ 9 $$-$$ 63m
$$ \Rightarrow $$ 50m $$=$$ $$-$$ 100 or 76m $$=$$ 82
$$ \Rightarrow $$ m $$=$$ $$ - {1 \over 2}$$
or m $$=$$ $${{41} \over {38}}$$
$$ \Rightarrow $$ y $$-$$ 1 $$=$$ $$-$$ $${1 \over 2}$$ (x $$-$$ 0)
or y $$-$$ 1 $$=$$ $${{41} \over {38}}$$ (x $$-$$ 0)
i.e x + 2y $$-$$ 2 $$=$$ 0
or 38y $$-$$ 38 $$-$$ 41x $$=$$ 0
$$ \Rightarrow $$ 41x $$-$$ 38y + 38 $$=$$ 0
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