A light beam emanating from the point A (3,10) reflects from the line 2 x + y -6=0 and then passes through the point B(5,6). The equation of the incident and reflected beams are respectively :

Question

A light beam emanating from the point A (3,10) reflects from the line 2 x + y -6=0 and then passes through the point B(5,6). The equation of the incident and reflected beams are respectively : (A) 4 x-3 y+18=0 and y=6 (B) x-2 y+8=0 and x=5 (C) x+2 y-8=0 and y=6 (D) none of these

Tardigrade Admin · Accepted Answer

slope of AB = -2 = slope of given line ⇒ PAB is isosceles . Equation of line MP is x -2y + 12 = 0 . Solving it with 2x + y -6 = 0 point P(0, 6) . Equation of AP is 4x -3y + 18 = 0 and equation of BP is y - 6 = 0

Q. A light beam emanating from the point $A (3, 10)$ reflects from the line $2 x + y - 6 = 0$ and then passes through the point $B (5, 6)$ . The equation of the incident and reflected beams are respectively :

$4 x - 3 y + 18 = 0$ and $y = 6$

$x - 2 y + 8 = 0$ and $x = 5$

$x + 2 y - 8 = 0$ and $y = 6$

none of these

slope of $A B = - 2 =$ slope of given line
$\Rightarrow P A B$ is isosceles . Equation of line MP is
$x - 2 y + 12 = 0$ . Solving it with $2 x + y - 6 = 0$
point P(0, 6) . Equation of AP is $4 x - 3 y + 18 = 0$
and equation of BP is $y - 6 = 0$

Q. A light beam emanating from the point A(3,10) reflects from the line 2x+y−6=0 and then passes through the point B(5,6). The equation of the incident and reflected beams are respectively :

4x−3y+18=0 and y=6

x−2y+8=0 and x=5

x+2y−8=0 and y=6