Home
Class 11
MATHS
Let P be a point on the circle x^2+y^2=9...

Let `P` be a point on the circle `x^2+y^2=9,Q` a point on the line `7x+y+3=0` , and the perpendicular bisector of `P Q` be the line `x-y+1=0` . Then the coordinates of `P` are `(0,-3)` (b) `(0,3)` `((72)/(25),(21)/(35))` (d) `(-(72)/(25),(21)/(25))`

Promotional Banner

Similar Questions

Explore conceptually related problems

Let P be a point on the circle x^(2)+y^(2)=9 , Q a point on the line 7x+y+3=0 , and the perpendicular bisector of PQ be the line x-y+1=0 . Then the coordinates of P are

Let P be a point on the circle x^2+y^2=9,Q a point on the line 7x+y+3=0 , and the perpendicular bisector of P Q be the line x-y+1=0 . Then the coordinates of P are (a) (0,-3) (b) (0,3) ((72)/(25),(21)/(35)) (d) (-(72)/(25),(21)/(25))

Let P be a point on the circle x^(2)+y^(2)=9,Q a point on the line 7x+y+3=0 ,and the perpendicular bisector of PQ be the line x-y+1=0. Then the coordinates of P are (0,-3)(b)(0,3)((72)/(25),(21)/(35))(d)(-(72)/(25),(21)/(25))

P is a point on the circle x^2+y^2=9 Q is a point on the line 7x+y+3=0 . The perpendicular bisector of PQ is x-y+1=0 . Then the coordinates of P are:

P is a point on the circle x^(2)+y^(2)=9Q is a point on the line 7x+y+3=0. The perpendicular bisector of PQ is x-y+1=0 Then the coordinates of P are:

P is a point on the parabola y^(2)=4x and Q is Is a point on the line 2x+y+4=0. If the line x-y+1=0 is the perpendicular bisector of PQ, then the co-ordinates of P can be:

P is a point on the straight line 7x-4y-29=0 . If the foot of the perpendicular drawn from P on the line 5x+2y+18=0 be N(-2,-4) , find the coordinates of P.

Find the reflection of the point Q(2,1) in the line y + 3 = 0