AP EAMCET202125 Aug 2021Evening ShiftMathematicsStraight LinesActual
Identify the point on the line 2x + 3y + 7 = 0, which is at a distance of + 3 units from (1, − 3).
Options
- A( 13 +9 13 , -3 13 +6 13 )
- B( 13 -9 13 , -3 13 -6 13 )
- C( 13 -9 13 , -3 13 +6 13 )
- D( 13 +9 13 , 3 13 -6 13 )
Correct answer
C. ( 13 -9 13 , -3 13 +6 13 )
Step-by-step solution
Let P( , ) be the point on the line 2x + 3y + 7 = 0 array rlrl & & 2 +3 +7 & =0 & & = ( -7-2 3 ) & & P & = ( , -7-2 3 ) array Given, point A = (1, − 3) aligned & A P=3 & ( -1)^2+ ( -7-2 3 +3 )^2 =3 & ^2+1-2 + 49+4 ^2+28 9 +9-14-4 =9 & 9 ^2+9-18 +49+4 ^2+28 -126-36 =0 aligned aligned & 13 ^2-26 -68=0 & 13 ^2-26 -68=0 & = 26 676+3536 26 & =1 9 13 = 13 9 13 aligned When, = 13 -9 13 , = -7-2 3 aligned & = -3 13 +6 13 P & = ( 13 -9 13 , -3 13 +6 13 ) aligned