AP EAMCET202319 May 2023Morning ShiftMathematicsStraight LinesActual
The point P (2,1) is translated to a point Q parallel to the line L x-y-4=0 by 2 3 units. If the point Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is
Options
- A2 x+2 y=1- 6
- Bx+y=3-3 6
- Cx+y=2- 6
- Dx+y=3-2 6
Correct answer
D. x+y=3-2 6
Step-by-step solution
Given L x-y-4=0 x (-4) + y (4) =1 Slope m₁ of line ' L ' is m₁=1 Since line ' L ' and line P Q is parallel to each other. Hence slope of line ' L ' = Slope of line P Q Hence equation of P Q can be written as aligned & y-1=m₁(x-2)=(1)(x-2) & y=(x-1) & Distance P Q=2 3 & (x-2)^2+(y-1)^2 =2 3 & (x-2)^2+(x-2)^2=12 & (x-2)^2=6 & (x-2)= 6 or x=2+ 6 or 2- 6 aligned Here x=2+ 6 is not valid because point Q lies in third quadrant. Hence x=2- 6 Therefore y=(2- 6 )-1=1- 6 Q(x, y)=Q(2- 6 , 1- 6 ) Let the slope of line perpendi