AP EAMCET201822 Apr 2018Morning ShiftMathematicsStraight LinesActual
The equation of the line passing through the point of intersection of the lines 2 x+y-4=0, x-3 y+5=0 and lying at a distance of 5 units from the origin, is
Options
- Ax-2 y-5=0
- Bx+2 y-5=0
- Cx+2 y+5=0
- Dx-2 y+5=0
Correct answer
B. x+2 y-5=0
Step-by-step solution
The equation of a line passing through the in ter section of 2x + y − 4 = 0 and x − 3y + 5 = 0 is (2x + y − 4) + λ(x − 3y + 5) = 0 …(i) ⇒ x(2 + λ) + y(1 − 3λ) + 5λ − 4 = 0 This is at a distance of 5 units from the origin. array ll & | 5 -4 (2+ )^2+(1-3 )^2 |= 5 & (5 -4)^2 4+ ^2+4 +1+9 ^2-6 =5 & (5 -4)^2 10 ^2-2 +5 =5 & 25 ^2+16-40 =50 ^2-10 +25 & 25 ^2+30 +9=0 array By solving, we get =- 3 5 Putting the value of =- 3 5 in Eq. (i), we get array rrrl (2 x+y-4)- 3 5 (x-3 y+5) =0 10 x+5 y-20-3 x+9 y-15 =0 7 x+14 y-35 =