The Coordinate Plane: Very Difficult Problems with Solutions

Solution:
The symmetric of point A with respect to point O is point A', which is found on the same line as A and O. O is halfway between A and A'. Point A is the vertex of a rectangle in quadrant I with a length of 2 and a width of 3 and whose diagonal is OA. Point A' is the vertex of a rectangle of the same size, but in quadrant III and with diagonal OA'. The coordinates of point A' are (-2, -3).

Solution:
AB is parallel to the x-axis and has a length of 7 - 2 = 5 units. The length of AM is equal to half of the length of AB = 2.5 units. The coordinates of point M are (2 + 2.5, 3) = (4.5, 3).

Solution:
If the square has an area of 9 square units, then its side has a length of 3 units. If O is where the axes meet, then points C and B are found on the x- and y-axes and they have the coordinates (3, 0) and (0, 3). Point A is at 3 units from the x-axis and at 3 units from the y-axis and it has the coordinates (3, 3).

Solution:
The points which are found on the circle must be at a distance of 2 units from O. The points which fulfill this condition are A, D, G and J. The others are at a distance greater than 2 from point O.

Solution:
Triangle ABC can be included inside square MOCN. We can express the area of the triangle as the difference between the area of the square and the area of the 3 right triangles. The area of square MOCN is 9 square units. The area of triangle AMB is 2 square units. The area of triangle BOC is 1.5 square units. The area of triangle CAN is 1.5 square units. Thus, the area of triangle ABC is 9 - (2 + 1.5 + 1.5) = 9 - 5 = 4 square units.