Show that the parallelogram formed by `ax+by+c=0, a_1 x+ b_1 y+c=0, ax+by+c_1 =0 and a_1 x+b_1 y+c_1 =0` will be a rhombus if `a^2 +b^2 = (a_1)^2 + (b_1)^2`.

the diagonals of the parallelogram formed by the the lines a_1x+b_1y+c_1=0 ,a_1x+b_1y+c_1 '=0 , a_2x+b_2y+c_1=0 , a_2x+b_2y+c_1 '=0 will be right angles if:

Prove that the area of the parallelogram formed by the lines a_1x+b_1y+c_1=0,a_1x+b_1y+d_1=0,a_2x+b_2y+c_2=0, a_2x+b_2y+d_2=0, is |((d_1-c_1)(d_2-c_2))/(a_1b_2-a_2b_1)| sq. units.

If the origin lies in the acute angle between the lines a_1 x + b_1 y + c_1 = 0 and a_2 x + b_2 y + c_2 = 0 , then show that (a_1 a_2 + b_1 b_2) c_1 c_2 lt0 .

If one of the roots of the equation ax^2 + bx + c = 0 be reciprocal of one of the a_1 x^2 + b_1 x + c_1 = 0 , then prove that (a a_1-c c_1)^2 =(bc_1-ab_1) (b_1c-a_1 b) .

Prove that the points (a ,\ b),\ (a_1,\ b_1) and (a-a_1,\ b-b_1) are collinear if a b_1=a_1b .

(1) The straight lines (2k+3) x + (2-k) y+3=0 , where k is a variable, pass through the fixed point (-3/7, -6/7) . (2) The family of lines a_1 x+ b_1 y+ c_1 + k (a_2 x + b_2 y + c_2) = 0 , where k is a variable, passes through the point of intersection of lines a_1 x + b_1 y + c_1 = 0 and a_2 x+ b_2 y + c_2 = 0 (A) Both 1 and 2 are true and 2 is the correct explanation of 1 (B) Both 1 and 2 are true and 2 is not a correct explanation of 1 (C) 1 is true but 2 is false (D) 1 is false but 2 is true

if a, b, c, a_1, b_1 and c_1 are non-zero complexnumbers satisfying a/a_1+b/b_1+c/c_1=1+i and a_1/a+b_1/b+c_1/c=0, where i=sqrt-1, the value of a^2/a_1^2+b^2/b_1^2+c^2/c_1^2 is (a)2i (b)2+2i (c)2 (d)None of these

If the points (a_1, b_1),\ \ (a_2, b_2) and (a_1+a_2,\ b_1+b_2) are collinear, show that a_1b_2=a_2b_1 .

Consider a system of linear equation in three variables x,y,z a_1x+b_1y+ c_1z = d_1 , a_2x+ b_2y+c_2z=d_2 , a_3x + b_3y + c_3z=d_3 The systems can be expressed by matrix equation [(a_1,b_1,c_1),(a_2,b_2,c_2),(c_1,c_2,c_3)][(x),(y),(z)]=[(d_1),(d_2),(d_3)] if A is non-singular matrix then the solution of above system can be found by X = A^(-1)B , the solution in this case is unique. if A is a singular matrix i.e. then the system will have no unique solution if no solution (i.e. it is inconsistent) if Where Adj A is the adjoint of the matrix A, which is obtained by taking transpose of the matrix obtained by replacing each element of matrix A with corresponding cofactors. Now consider the following matrix. A=[(a,1,0),(1,b,d),(1,b,c)], B=[(a,1,1),(0,d,c),(f,g,h)], U=[(f),(g),(h)], V=[(a^2),(0),(0)], X=[(x),(y),(z)] If AX=U has infinitely many solutions then the equation BX=U is consistent if

Show that the circle passing through the origin and cutting the circles x^2+y^2-2a_1x-2b_1y+c_1=0 and x^2+y^2-2a_2x -2b_2y+c_2=0 orthogonally is |(x^2+y^2,x,y),(c_1,a_1,b_1),(c_2,a_2,b_2)|=0