The image of `P(a ,b)` on the line `y=-x` is `Q` and the image of `Q` on the line `y=x` is `R` . Then the midpoint of `P R` is (a) `(a+b ,b+a)` (b) `((a+b)/2,(b+2)/2)` (c) `(a-b ,b-a)` (d) `(0,0)`

The image of P(a,b) on the line y=-x is Q and the image of Q on the line y=x is R . then the midpoint of PR is

The image of P(a,b) on the line y=-x is Q and the image of Q on the line y=x is 'R' find the mid-point of 'P' and 'R'

The image of P(a,b) on the line y=-x is Q and the image of Q on the line y=x is R find the mid-point of P and R

The image of the point P(3,5) w.r.t the line y=x is the point Q and the image of Q along the line y=0 is the point R(a, b), then (a, b)=

The image of the point P(3,5) with respect to the line y=x is the point Q and the image of Q with respect to the line y=0 is the point R(a,b), then (a,b) is equal to:

If R(x,y) is a point on the line segment joining the points P(a,b) and Q(b,a), then prove that x+y=a+b

If R(x,y) is a point on the line segment joining points P(a,b) and Q(b,a) , then prove that x + y = a + b . (a ne b)

If (x_1, y_1)&(x_2, y_2) are the ends of diameter of a circle such that x_1&x_2 are the roots of the equation a x^2+b x+c=0 and y_1&y_2 are the roots of the equation p y^2+q y+c=0 . Then the coordinates f the centre of the circle is: (b/(2a), q/(2p)) (b) (-b/(2a),=q/(2p)) (b/a , q/p) (d) None of these

Locus of the image of a point A(0,0) in the line (2a+b)x+(3a-2b)y+(b-5a)=0 is (where a and b are parameters).

Line a x+b y+p=0 makes angle pi/4 with xcosalpha+ysinalpha=p ,p in R^+ . If these lines and the line xsinalpha-ycosalpha=0 are concurrent, then (a) a^2+b^2=1 (b) a^2+b^2=2 (c) 2(a^2+b^2)=1 (d) none of these