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.

Solution:
From [tex]D(-4, -1)[/tex] to [tex]B(6, 4)[/tex] the abscissa increases by [tex]10[/tex] and the ordinate by [tex]5[/tex], so along [tex]DB[/tex] the ordinate grows half as fast as the abscissa.
From [tex]D[/tex] to the [tex]y[/tex]-axis the abscissa increases by [tex]4[/tex] (from [tex]-4[/tex] to [tex]0[/tex]), so the ordinate increases by [tex]2[/tex]: [tex]M(0, -1+2)=M(0, 1)[/tex].
[tex]a=AM=1-(-2)=3[/tex], [tex]b=CM=6-1=5[/tex]
[tex]a:b=3:5[/tex]

Solution:
The right angle can be at [tex]O[/tex], at [tex]M[/tex] or at [tex]N[/tex].
Right angle at [tex]O[/tex] ([tex]ON=OM[/tex], [tex]ON\perp OM[/tex]): [tex]N(-2, 2)[/tex] or [tex]N(2, -2)[/tex].
Right angle at [tex]M[/tex] ([tex]MN=MO[/tex], [tex]MN\perp MO[/tex]): [tex]N(4, 0)[/tex] or [tex]N(0, 4)[/tex].
Right angle at [tex]N[/tex] ([tex]OM[/tex] is the hypotenuse): [tex]N(2, 0)[/tex] or [tex]N(0, 2)[/tex].
In total [tex]2+2+2=6[/tex] possibilities.

Solution:
Area of [tex]ABC[/tex]: enclose it in the rectangle [tex]-5\le x\le 2[/tex], [tex]-6\le y\le 0[/tex] with area [tex]7\times 6=42[/tex] and subtract three right triangles:
[tex]S_{ABC}=42-\frac{2\times 6}{2}-\frac{7\times 2}{2}-\frac{5\times 4}{2}=42-6-7-10=19[/tex]
Triangles [tex]APC[/tex] and [tex]ABC[/tex] have the same height from [tex]C[/tex], so [tex]S_{APC}=\frac{AP}{AB}\times S_{ABC}=\frac{2}{5}\times 19[/tex].
Triangles [tex]APM[/tex] and [tex]APC[/tex] have the same height from [tex]P[/tex], so
[tex]S_{APM}=\frac{AM}{AC}\times S_{APC}=\frac{4}{5}\times \frac{2}{5}\times 19=\frac{152}{25}=6. 08[/tex]