Answer:
C. 
Step-by-step explanation:
c
Set up the given triangle on x-y coordinates with right angle at (0,0). So the two vertices are at (5,0) and (0,2
![sqrt{x} n]{3}](https://tex.z-dn.net/?f=sqrt%7Bx%7D%20n%5D%7B3%7D%20)
)
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>

for a pt (x,y) on line sx+ty=1, the minimum of

equals to

smallest value happens at

so area is

hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145
Yes the answer would be B
15: M= 2
B= -5
Slope intercept form= y=2x-5
16: y=5/8x + 1/2