7quad (iv)(x)/(2)+(2y)/(3)=-1" str "x-(y)/(3)=3

Find the S.D. between the lines : (i) (x)/(2) = (y)/(-3) = (z)/(1) and (x -2)/(3) = (y - 1)/(-5) = (z + 4)/(2) (ii) (x -1)/(2) = (y - 2)/(3) = (z - 3)/(2) and (x + 1)/(3) = (y - 1)/(2) = (z - 1)/(5) (iii) (x + 1)/(7) = (y + 1)/(-6) = (z + 1)/(1) and (x -3)/(1) = (y -5)/(-2) = (z - 7)/(1) (iv) (x - 3)/(3) = (y - 8)/(-1) = (z-3)/(1) and (x + 3)/(-3) = (y +7)/(2) = (z -6)/(4) .

Find each of the following products: (i) (x + 3) (x - 3) (ii) (2x + 5)(2x - 5) (ii) (8 + x)(8 - x) (iv) (7x + 11y) (7x - 11y) (v) (5x^(2) + (3)/(4) y^(2)) (5x^(2) - (3)/(4) y^(2)) (vi) ((4x)/(5) - (5y)/(3)) ((4x)/(5) + (5y)/(3)) (vii) (x + (1)/(x)) (x - (1)/(x)) (viii) ((1)/(x) + (1)/(y)) ((1)/(x) - (1)/(y)) (ix) (2a + (3)/(b)) (2a - (3)/(b))

Simplify : (i) (5x - 9y) - (-7x + y) (ii) (x^(2) -x) -(1)/(2)(x - 3 + 3x^(2)) (iii) [7 - 2x + 5y - (x -y)]-(5x + 3y -7) (iv) ((1)/(3)y^(2) - (4)/(7)y + 5) - ((2)/(7)y - (2)/(3)y^(2) + 2) - ((1)/(7)y - 3 + 2y^(2))

Evaluate : (iv) (3x - (1)/(2y) ) (3x + (1)/( 2y))

Write the following squares of binomials as trinomials: (x+(x^(2))/(2))^(2) (ii) ((x)/(4)-(y)/(3))(3x-(1)/(3x))^(2)( iv )((x)/(y)-(y)/(x))^(2)

The angles between the lines (x+1)/(-3) =(y-3)/(2)=(z+2)/(1) and x=(y-7)/(-3) =(z+7)/(2) are

Find the following products: (x+2)(x+3) (ii) (x+7)(x-2)y-4)(y-3)( iv )(y-7)(y+3)(2x-3)(2x+5)( vi )(3x+4)(3x-5)

Factorise : (iv) (2x - y)^(3) - (2x - y)

Find the following products: (i) (x+2)(x+3) (ii) (x+7)(x-2) (iii) y-4)(y-3) (iv) (y-7)(y+3) (vi) (2x-3)(2x+5) (vii) (3x+4)(3x-5)