MHT CET202416 May 2024Evening ShiftMathematicsPair of LinesActual
If P₁ and P₂ are perpendicular distances (in units) from point (2,-1) to the pair of lines 2 x^2-5 x y+2 y^2=0 , then the value of P ₁ P ₂ is
Options
- A2
- B5
- C10
- D4
Correct answer
D. 4
Step-by-step solution
array ll & Given equation of pair of lines is & 2 x^2-5 x y+2 y^2=0 & 2 x^2-4 x y-x y+2 y^2=0 & 2 x(x-2 y)-y(x-2 y)=0 & (2 x-y)(x-2 y)=0 array separate equations of the lines are 2 x-y=0 and x-2 y=0 Perpendicular distances of the above lines from (2,-1) are aligned & P₁= | 2(2)-(-1) (2)^2+(-1)^2 |= | 5 5 | and & P₂= | 2-2(-1) (1)^2+(-2)^2 |= | 4 5 | aligned P₁ P₂= 5 5 4 5 =4