Given `A-=(1,1)` and `A B` is any line through it cutting the x-axis at `Bdot` If `A C` is perpendicular to `A B` and meets the y-axis in `C` , then the equation of the locus of midpoint `P` of `B C` is (a) `x+y=1` (b) `x+y=2` (c) `x+y=2x y` (d) `2x+2y=1`

A straight line through the point (1,1) meets the X-axis at A and Y-axis at B. The locus of the mid-point of AB is

Given the graph of y=f(x) . Draw the graphs of the followin. (a) y=f(1-x) (b) y=-2f(x) (c) y=f(2x) (d) y=1-f(x)

The locus of the center of the circle touching the line 2x-y=1 at (1,1) is (a)x+3y=2 (b) x+2y=2( c) x+y=2 (d) none of these

Let A B C be a triangle. Let A be the point (1,2),y=x be the perpendicular bisector of A B , and x-2y+1=0 be the angle bisector of /_C . If the equation of B C is given by a x+b y-5=0 , then the value of a+b is (a)1 (b) 2 (c) 3 (d) 4

Given ABCD is a parallelogram wilth vertices as A(1,2),B(3,4) and C(2,3) . Then the equation of line AD is (A) x+y+1=0 (B) x-y+1=0 (C) -x+y+1=0 (D) x+y-1=0

A line intersects x-axis at A(2, 0) and y-axis at B(0, 4) . A variable lines PQ which is perpendicular to AB intersects x-axis at P and y-axis at Q . AQ and BP intersect at R . Image of the locus of R in the line y = - x is : (A) x^2 + y^2 - 2x + 4y = 0 (B) x^2 + y^2 + 2x + 4y = 0 (C) x^2 + y^2 - 4y = 0 (D) x^2 + y^2 + 2x - 4y = 0