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**deej813** ABCD is a trapezium with corners A(6,0), B(0,4), C(-2,0) and D(1,-2). The line passing through AB has the equation 2x + 3y - 12 = 0

a. Find acute angle between AB and the pos. x-axis

Is there a formula for this?? I can't remember it

b. Midpoint AB (M)

that's just [(6+0)/2 , (0+4)/2]

which is (3,2)

is that right??

c. show equation of line CD is 2x + 3y + 4 = 0

m=-2/3 x=-2 y=0

using y-y1=m(x-x1)

y-0=-2/3(x+2)

3y=-2x-4

then 2x + 3y + 4 = 0

yes???

d. show distance from M to CD is (16sqrt13)/13

do you just use the perpendicular distance formula??

so that would be

|ax1 + by1 + c|/(sqrta^2 + b^2)

and you get 16/sqrt13

so you just rationalise and get (16sqrt13)/13

e. gradient CD = -2/3

m AB = (0-4)/(6-0)

= -4/6

=-2/3

therefore parallel because m1 = m2

f.given CD = 5 units, calculate distance AB and find area of trapezium

dAB = sqrt[(x2-x1)^2 + (y2-y1)^2]

= sqrt52

and area = 1/2(a+b)h

= 1/2 x (5+sqrt52) x [(16sqrt13)/13]

so you get

=27.09units^2

f. M is centre of circle, CD is tangent, what is the equation??

M(3,2) CD- 2x + 3y + 4 = 0

is it just

(x-a)^2 + (y-b)^2 = r^2

which is (3-2)^2 + (2-3)^2 = ((16sqrt13)/13)^2