Show a quadrilateral ABCD in A is(2,8) and B is (8,6) .the point c lies on the perpendicular bisector of AB and the point D lies on the y- axis. the equation of BC is 3y=4x-14 and angle DAB=90° find (a) the equation of AD
(b) the coordinates of D
(c) the equation of perpendicular bisector of AB
(d) the coordinates of C
show the area of triangle ABC is 10 units and find the area of the quadrilateral ABCD

(a) y =  3x + 2; (b) (0,2); (c) y = 3x - 8; (d) (2,-2)

(e) Area of ∆ABC = 30; (f) Area of ABCD = 40

Step-by-step explanation:

(a) Equation of AD

(i) Slope of AB

m₁ = (y₂ - y₁)/(x₂ - x₁) = (6 - 8)/(8 - 2)= -2/6 = -⅓

(ii) Slope of AD

m₂ = 3

(ii) y-intercept

y = 3x + b

8 = 3(2) + b= 6 + b

b = 2

The equation of AD is y = 3x + 2.

(b) Coordinates of D

The coordinates of D are (0,2).

(c) Equation of perpendicular bisector of AB

(i) Mid-points of AB

x = ½(x₂ + x₁) = ½(8 + 2) = ½(14) = 5

y = ½(y₂ + y₁) = ½(6 + 8) = ½(14) = 7

The coordinates of the mid-point are (5,7).

Slope = 3

y = mx + b

7 = 3(5) + b = 15 + b

b = 7 - 15 = -8

The equation of the perpendicular bisector is y = 3x - 8.

(d) Coordinates of C

C is at the intersection of BC and the perpendicular bisector of AD.

 y = 3x -    8

3y = 4x - 14

3y = 9x - 24

 0 = 5x - 10

5x = 10

 x =   2

 y = 3(2) - 8  = 6 - 8 = -2

The coordinates of C are (2,-2).

(e) Area of ∆ABC

A = ½bh = ½ × 10 × 6 = 30

(f) Area of ABCD

Area of ∆ACD = ½bh = ½ × 10 × 2 = 10

Area of ABCD = ∆ACD + ∆ABC =10 + 30 = 40

dude   i don't even know. i've been trying for hours.

step-by-step explanation:

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: