The area of the parallelogram formed by the lines 3x + 4y = 7a; 3x + 4y = 7b; 4x + 3y = 7c and 4x + 3y = 7d is- /(a-b)(c-d)| (B)|(a - b) (c-d)| |(a-b)(c-d)) 49 (D) 7|(a - b) (c-d)|

If b gt a , d gt c then are of quadrilateral formed by straight lines x = b, x = a, y = c and y =d is

The area of the parallelogram formed by the lines 3x-4y + 1=0, 3x-4y +3=0,4x-3y-1=0 and 4x -3y -2 =0, is (A) 1/7 sq units (B) 2/7 sq units (C) 3/7 sq units (D) 4/7 sq units

The area of the parallelogram formed by the lines 3x-4y+1=0,3x-4y+3=0,4x-3y-1=0 and 4x-3y-2=0, is (A) (1)/(7) squnits (B) (2)/(7) squnits (C) (3)/(7) squnits (D) (4)/(7) squnits

A : The area of the parallelogram formed by 4x-7y-13=0, 8x-y-39=0, 4x-7y+39=0, 8x-y+13=0 is 52. R : The area of the parallelogram formed by a_(1)x+b_(1)y+c_(1)=0, a_(1)x+b_(1)y+d_(1)=0, a_(2)x+b_(2)y+c_(2)=0, a_(2)x+b_(2)y+d_(2)=0 is |((c_(1)-d_(1))(c_(2)-d_(2)))/(a_(1)b_(2)-a_(2)b_(1))|

If a : b = c : d , prove that : (6a + 7b) (3c - 4d) = (6c + 7d) (3a - 4b) .

If (7a + 8b)/( 7c + 8d) = (7a - 8b)/(7c - 8d), prove that : (i) a:b=c:d (ii) (4a - 5b)/(4c - 5d)= (3a + 4b)/(3c + 4d)

If A:B=2:3 , B:C = 4:5, C:D = 6:7. then A : B : C : D is-

If a: b= 2: 3, b: c= 4: 5 and c: d= 6: 7 , find : a: b: c:d .

If 7x + 3y + 9 = 0 and y = kx + 7 are two parallel lines than k is (a)3/7 (b)-7/3 (c)3 (d)7

The points on the line x+y =4 lying at a unit distance from the line 4x + 3y - 10 =0 are a) (-7, 11) , (3,1) b) (7, -11), (3,-1) c) (-7, 11) , (-3,7) d) (7,-3), (11, -7)