The diagonals AC and BD of a rhombus ABCD intersect at the point (3,4) . If BD =2 2 , ~A =(1,2), B =( , ) , D=( , ) and < < < , then + - =

Step-by-step solution

B D=2 2 O B=O D= 2 (3,4) is mid point of A C C=(5,6) So, O A=O C= 4+4 =2 2 In A O B, O A^2+O B^2=A B^2 A B^2=10 A B= 10 A B=B C=C D=D A= 10 Since O(3,4) is mid point of B D , so aligned & + =6 and + =8 & C D= 10 C D^2=10 & ( -5)^2+( -6)^2=10 & (1- )^2+( -6)^2=10 & ( -6)^2=10-(1- )^2...(i) aligned Also, O B^2=( 2 )^2 ( -3)^2+( -4)^2=2 ( -4)^2=2-( -3)^2...(ii) Equation (ii) - (i) aligned & ( -4)^2-( -6)^2=2-( -3)^2-10+(1- )^2 & ( + -10)( - +2)=4 -16 & (8-10)( - +2)=4 -16 & - +2=-2 +8 & - =-2 +6 aligned Now, + - =( -