f4- amath

In triangle ABC, the vertex B is (-4,-1). Angle BAC is bisected by the line x=0. The equation of the altitude AD on the side BC is 6x-y+3=0. Find
(a) the coordinates of A
Vertex A is the intersection of the altitude AD and the line x=0 (i.e. y-axis).
The equation of the altitude AD is :
6x-y+3=0 . . . . . (1)
Put x=0 into (1) :
-y+3=0
y=3
The coordinates of A are (3,0)

(b) the equation of BC
From (1) : y=6x-3
Slope of AD is : 6
Slope of BC x Slope of AD=-1
Slope of BC is : -1/6
The equation of BC is : y-(-1)=(-1/6)[x-(-4)]
y+1=(-1/6)(x+4)
6y+6=-x-4
x+6y+10=0

(c) the coordinates of C
Slope of AB is : [3-(-1)]/[0-(-4)]=4/4=1
Since angle BAC is bisected by the line x=0 (i.e. the y-axis),
Slope of AC is : -1
The equation of AC is :
y-3=-1(x-0)
y-3=-x
x+y-3=0 . . . . . (2)
The equation of BC is :
x+6y+10=0 . . . . (3)
(3)-(2) : 5y+13=0
y=-13/5
x=28/5
The coordinates of C are (28/5,-13/5)

(d) the ratio that AC is divided by the x-axis
The equation of AC is :
x+y-3=0
When y=0
x-3=0
x=3
AC cuts the x-axis at the point E(3,0).
Let the ratio be r (i.e. AE / EC = r).
0=[3+(-13/5)r]/(1+r)
3+(-13/5)r=0
15-13r=0
r=15/13
The ratio that AC is divided by the x-axis is 15 : 13.