1. ## trapezium

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

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

2. Originally Posted by 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

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

a- 2x + 3y - 12 = 0
to find the acute angel between the line and the positive x axis there is many ways the way i9 remmber is to find the deravative of this equation
so by arranging this equation y=4-2/3 x
so dy/dx = -2/3
so the tan of the angle u want is -2/3 by using calculator
and i think the angle u want will be 142.56