Question

The vertices of a family of triangles have integer coordinates. If two of the vertices of all the triangles are (0, 0) and (6, 8), then the least value of areas of the triangles is

• A. $ 1 $
• B. $ \frac{3}{2} $
• C. $ 2 $
• D. $ \frac{5}{2} $
• E. $ 3 $

Answer 1

Answer: (A)

Let the third vertex be (a, b).
$ \therefore $ Area of $ \Delta =\frac{1}{2}\left| \left| \begin{matrix} 0 & 0 & 1 \\ a & b & 1 \\ 6 & 8 & 1 \\ \end{matrix} \right| \right| $
$=\frac{1}{2}|[8a-6b]| $ As (a, b) are integers, so we take (0, 0), (1, 1), (1, 2) At (0, 0), $ \Delta =0, $ it is not possible. At (1, 1), $ \Delta =1 $ At (1, 2), $ \Delta =2 $ Here, we see that minimum area is 1.