The matrix `[(0,1),(1,0)]` is the matrix of reflection in the line (A) `x-y=0` (B) `x+y=0` (C) `x-y=1 (D) `x+y=1`

the matrix [(0,1),(1,0)] is the matrix reflection in the line

The matrix of the transformation reflection in the line x+y=0 is (A) [(-1,0),(0,-1)] (B) [(1,0),(0,-1)] (C) [(0,1),(1,0)] (D) [(0,-1),(-1,0)]

x=2, y=−1 is a solution of the linear equation (a) x+2y=0 (b) x+2y=4 (c) 2x+y=0 (d) 2x+y=5

Consider the parabola P touching x-axis at (1,0) and y-axis (0,2). Directrix of parabola P is : (A) x-2y=0 (B) x+2y=0 (C) 2x-y=0 (D) 2x+y=0

Given ABCD is a parallelogram with 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 image of line 2x+y=1 in line x+y+2=0 is : (A) x+2y-7=0 (B) 2x+y-7=0 (C) x+2y+7=0 (D) 2x+y+7=0

The following line(s) is (are) tangent to the curve y=x^2 - x (A) x-y=0 (B) x+y=0 (C) x-y=1 (D) x+y=1

The normal at the point (1,1) on the curve 2y+x^2=3 is (A) x - y = 0 (B) x y = 0 (C) x + y +1 = 0 (D) x y = 0

Consider point P(x, y) in first quadrant. Its reflection about x-axis is Q(x_(1), y_(1)) . So, x_(1)=x and y(1)=-y . This may be written as : {(x_(1)=1. x+0.y),(y_(1)=0. x+(-1)y):} This system of equations can be put in the matrix as : [(x_(1)),(y_(1))]=[(1,0),(0,-1)][(x),(y)] Here, matrix [(1,0),(0,-1)] is the matrix of reflection about x-axis. Then find the matrix of reflection about the line y=x .

The normal at the point (1,1) on the curve 2y+x^2=3 is (A) x + y = 0 (B) x y = 0 (C) x + y +1 = 0 (D) x y = 0