It is not recommended that points be marked with X, let's marked with C(6,6)=(Xc,Yc)
The coordinates of the point C(Xc,Yc) which belongs to the line AB and divides line AB in a ratio m : n = 1 : 2 or m/n=1/2 are get it with following formula
Xc=(Xa+(m/n)Xb) / (1+(m/n)) and Yc=(Ya+(m/n)Yb) / (1+(m/n))
We have A(2,2)=(Xa,Ya) and B(14,14)=(Xb,Yb)
When we replace given coordinates we get
Xc=(2+(1/2)*14) / (1+(1/2)) = (2+7) /(3/2) = 9/(3/2) = (9*2)/3 = 3*2 =6 => Xc=6
Yc=(2+(1/2)*14) / (1+(1/2)) = (2+7) / (3/2) = 9/(3/2) = (9*2)/3 = 3*2 =6 => Yc=6
C(Xc,Yc)=(6,6)
Good luck!!!
The x intercept of -5y=4-2x equals (2,0)
To get the midsegment, namely HN, well, we need H and N
hmm so.... notice the picture you have there, is just an "isosceles trapezoid", namely, it has two equal sides, the left and right one, namely JL and KM
the midpoint of JL is H and the midpoint of KM is N
thus


![\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ 4q}}\quad ,&{{ 4n}})\quad % (c,d) M&({{ 4p}}\quad ,&{{ 0}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{4p+4q}{2}\quad ,\quad \cfrac{0+4n}{2} \right) \\\\\\ \left( \cfrac{2(2p+2q)}{2},\cfrac{4n}{2} \right)\implies \boxed{[(2p+2q), 2n]\impliedby N}\\\\ -----------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AK%26%28%7B%7B%204q%7D%7D%5Cquad%20%2C%26%7B%7B%204n%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AM%26%28%7B%7B%204p%7D%7D%5Cquad%20%2C%26%7B%7B%200%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%0A%25%20%20%20coordinates%20of%20midpoint%20%0A%5Cleft%28%5Ccfrac%7B4p%2B4q%7D%7B2%7D%5Cquad%20%2C%5Cquad%20%5Ccfrac%7B0%2B4n%7D%7B2%7D%20%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft%28%20%5Ccfrac%7B2%282p%2B2q%29%7D%7B2%7D%2C%5Ccfrac%7B4n%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cboxed%7B%5B%282p%2B2q%29%2C%202n%5D%5Cimpliedby%20N%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C)
√(-4-12)^2+(1-1)^2= √(-16)^2+0= √16^2=16
Answer is the last option: 16 units