1. m is the midpoint of ln and o is the midpoint of np.
solve for x, given mo = 2x +6 and lp = 8x – 20.

2. in △bec, f is the centroid and ac = 12. find a and f and fc. in △bec, f is the centroid and ac = 12. find a and f and fc.

3. da bisects angle bac. find dc.

4. abc has the vertices a(0, -1), b( 4, -1) and c (1, -4). sketch a graph of apc and use it to find the orthocenter of abc. then list the steps you took to find the orthocenter, including any necessary points or slopes you had to derive.

5. what are the properties of the following:
a. circumcenter:
b. incenter:
c. centroid:
d. orthocenter:

1. M is the midpoint of LN and O is the midpoint of NP. This makes the triangle MNO equal to half of LNP. Then you can get this equation
MO= (1/2) LP

If you insert MO = 2x +6 and LP = 8x – 20 the calculation would be:
2x+6= (1/2)( 8x-20)
2x+6= 4x-10
2x-4x= -10 - 6
-2x= -16
x=8

2. Centroid is the point that intersects with three median lines of the triangle. The centroid should divide the median lines into 1:2 ratio. In AC lines, A located in the base so A.F:FC would be 1:2

Then, the answer would be:
A.F= 1/(1+2) * AC
A.F= 1/3 * 12= 4

FC= 2/(1+2) * AC
FC= 2/3 * 12= 8

3. Since
∠BAD=∠DAC
∠ABD=∠ACD
AD=AD
The triangle ABD and ACD are similar. You can get this equation
BD=DC
x+8= 3x+12
x-3x= 12-8
-2x=4
x=-2

DC=3x+12= 3(-2) +12= 6

4. Orthocenter made by intersection of triangle altitude
A
BC lines slope would be (-4)-(-1)/1-4= -3/-3= 1. The altitude line slope would be -1, the function would be:
y=-x +a
0= 1+a
a=-1
y=-x-1
B
AC lines slope would be (-4)-(-1)/1-0= -3. The altitude line slope would be 1/3, the function would be:
y=1/3x+a
-1=1/3(4)+a
a=-7/3
y=1/3x - 7/3

C
BC lines slope would be (-1)-(-1)/4 = 0/4. 
The line would be 
0=x+a
a=-1
0=x-1
x=1

y=-x-1 = 1/3x-7/3
-x-(1/3x)=-7/3 +1
-4/3x= -4/3
x=1

y=-x-1
y=-1-1= -2
The orthocenter would be (1,-2)

5. 
a.Circumcenter: the intersection of perpendicular bisector lines
b.Incenter: the intersection of bisector lines
c.Centroid: the intersection of median lines
d.Orthocenter: the intersection of altitude lines