Problem:
Two lines with slopes 21​ and 2 intersect at (2,2). What is the area of the triangle enclosed by these two lines and the line x+y=10?
Answer Choices:
A. 4
B. 42​
C. 6
D. 8
E. 6 \sqrt
Solution:
Let P(2,2) be the intersection point. The two lines have equations y=21​x+1 and y=2x−2. They intersect x+y=10 at A(6,4) and B(4,6). Consider AB to be the base of the triangle; then the altitude of the triangle is the segment joining (2,2) and (5,5). By the Distance Formula, the area of △PAB is
Note: The area of the triangle with vertices (2,2),(6,4), and (4,6) can be calculated in a number of other ways, such as by enclosing it in a 4×4 square with sides parallel to the coordinate axes and subtracting the areas of three right triangles; by splitting it into two triangles with the line y=4; by the shoelace formula:
or by observing that there are 4 lattice points in the interior of the triangle and 6 lattice points on the boundary, and using Pick's Formula: 4+26​−1=(C)6​.
