(1)(1) Finding B.
Slope of AEAE.
(6-4)/(-1-3)=-1/26−4−1−3=−12
Since AEAE is perpendicular to EBEB, the slope of BEBE is 22 (negative reciprocal of -1/2−12).
The (x, y)(x,y) in here will be B's coordinates.
"Slope of BE"(2)=(y-4)/(x-3)Slope of BE(2)=y−4x−3
2x-6=y-42x−6=y−4
2x-y=22x−y=2
"Slope of AB"(1/3)=(y-6)/(x+1)Slope of AB(13)=y−6x+1
x+1=3y-18x+1=3y−18
x-3y=-19x−3y=−19
Solve the system of equations:
2x-y=22x−y=2
-2x+6y=38−2x+6y=38
5y=405y=40
y=8,x=5y=8,x=5
Thus B(5,8)(5,8).
(2)(2)
The height of Triangle EBCEBC is the difference of yy coordinates of BB and EE.
Which is 8-4=48−4=4
The base of the Triangle is the difference of the xx coordinates of EE and CC
(The xx coordinate of CC is unkown.)
Which is (x-3)(x−3).
Use the area of triangle area of EBCEBC.
EBC=1/2(x-3)(4)EBC=12(x−3)(4)
24=2x-624=2x−6
x=15x=15
Thus C(15,4)(15,4)
(3)(3) Finding D.
The slope of AE is the same as AD.
Thus:
AE=(y-4)/(x-3)AE=y−4x−3
-1/2=(y-4)/(x-3)−12=y−4x−3
But the xx coordinate of C is the same as D.
-1/2=(y-4)/(15-12)−12=y−415−12
-3=2y-8−3=2y−8
y=5/2y=52
Thus: D(15,5/2)(15,52)
