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`

The vertices of a parallelogram in order are A(1,2), B(4, y), C(x, 6) and D(3,5). Then (x, y) is

The vertices of a parallelogram in order are A(1,2), B (4,y), C (x,6) and D(3,5) . Then (x,y) is :

The vertices of a parallelogram in order are A(1, 2), B(4, y), C(x, 6), D(3, 5), then (x, y) =

Let ABC is a triangle whose vertices are A(1, –1), B(0, 2), C(x', y') and area of triangleABC is 5 and C(x', y') lie on 3x + y – 4lambda = 0 , then

If (2, -2),(-1,2), (3,5) are the vertices of a triangle then the equation of the side not passing through (2,-2) is (A) x + 2y +1 = 0 (B) 2x - y- 4= 0 (C) X-3y -3= 0 (D) 3x – 4y + 11 = 0

The equation of the tangent to the curve y=sec x at the point (0,1) is (a) y=0 (b) x=0 (c) y=1 (d) x=1

In a triangle ABC ,the bisectors of angles BandC lies along the lines x=y and y=0 . If A is (1,2) ,then the equation of line BC is (a) 2x+y=1 (b) 3x-y=5 (c) x-2y=3 (d) x+3y=1

Q. the equation of image of the pair of lines y=|2x-1| in y-axis is (A) 4x^2-y^2-4x+1=0 (B) 4x^2-y^2+4x+1=0 (C) 4x^2+y^2+4x+1=0 (D) 4x^2+y^2-4x+1=0

3x - 2y + 1 = 0 and 2x - y = 0 are the equation of the sides AB and AD of the parallelogram ABCD and the equation of a diagonal of the parallelogram is 5x - 3y - 1 = 0 . The equation of the other diagonal of the parallelogram is : (A) x-y+1=0 (B) x-y-1=0 (C) 3x+5y+13=0 (D) 3x+5y=13

Find the equation of the tangent line to the curve y=(3x^2-1)/x at the point (1,2) is (a) 4x-y-1=0 (b) 4x+y+2=0 (c)x-y-1=0 (d) 4x-y-2=0