Problem: ABC is a triangle: A=(0,0),B=(36,15) and both the coordinates of C are integers. What is the minimum area â–³ABC can have?
Answer Choices:
A. 21​
B. 1
C. 23​
D. 213​
E. there is no minimum
Solution:
Since A=(0,0) and B=(36,15), we know that base AB of △ABC has length 3122+52​=39. We must choose C so that the height of △ABC is minimum. The height is the distance to C=(x0​,y0​) from the line AB. This line is 5x−12y=0. In general, the distance from (x0​,y0​) to the line ax+by=c is
a2+b2​∣ax0​+by0​−c∣​.
So in this case the distance is ∣5x0​−12y0​∣/13. Since x0​ and y0​ are integers, the smallest this expression could be is 1/13. (0/13 is not possible, for then C would be on line AB and we would not have a triangle.) The value 1/13 is achieved, for instance, with C=(5,2) or (7,3). Thus the minimum area is (1/2)bh=(1/2)(39)(1/13)=3/2.
OR
The triangle with vertices (x1​,y1​),(x2​,y2​),(x3​,y3​) has area
Setting (x1​,y1​)=A,(x2​,y2​)=B,(x3​,y3​)=C, we have
Area ABC=21​ abs ∣∣∣∣∣​36x3​​15y3​​∣∣∣∣∣​=21​∣36y3​−15x3​∣=23​∣12y3​−5x3​∣
If x3​,y3​ are integers, the least nonzero value ∣12y3​−5x3​∣ could have is 1. Indeed, when x3​=5,y3​=2, it is 1. So 3/2 is the minimum area.