Solution to Let A,B,C єR2 where A = { (x,y) / y = 2x + 1} , … - Sikademy
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Archangel Macsika

Let A,B,C єR2 where A = { (x,y) / y = 2x + 1} , B = { (x,y) / y = 3x} and C = { (x ,y) / x - y = 7} . Determine each of the following: i. A intersection ii.( B intersection C ) complement iii. A complent union C complent iv. B complent union C complent

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Note that A, B and C just represent lines in 2-D as shown:

The event (A\cap B) represents the set of points that lie on both A and B. Clearly, this is their point of intersection, i.e., the singleton set {(1,3)}.

Similarly, the set (B\cap C) stands for those points which are common to both the lines, which is again the singleton set {(-3.5, -10.5)}.

Now, using DeMorgan's laws, one can rewrite (iii) as (\overline{\overline{A}\cup\overline{C}} = \overline{\overline{A}}\cap\overline{\overline{C}} = A\cap C)

Hence, this set is just the singleton set {(-8,-15)}, where (-8,-15) is the point of intersection of the 2 lines represented by A and C .

Now, by DeMorgan's laws, one can rewrite (iv) as (\overline{B}\cup\overline{C} = \overline{B\cap C}) which is just the complement of set {(-8,-15)}. Hence the complement of this set will be (\mathbb{R}^2\setminus \{(-8,-15)\}) .Hence the answer to all the parts are as follows: (\mathbb{R}^2\setminus\{(-8,-15)\})

{(1,3)}

{(-8,-15)\}

{(-3.5, -10.5)}

(\mathbb{R}^2\setminus\{(-8,-15)\})

i. A intersection B = { (1,3)}

ii. B intersection C = { (4/7, 12/7)}

iii. {(x,y)|y=2x+1 or y= x-7}

iv. { (x,y) | y not equal = 3x and x-y not equal = 7}


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