Answer:
Part A: We have the points A = ( 1,3 ) , B = ( 3,1 ) and C = ( 3,-3 ).
We will first find the equation of line AB and AC.
Now, the slope of AB is .
So, substituting in y = mx + b along with the point ( 1,3 ) gives,
3 = -1 + b i.e. b = 4.
So, the equation of AB is y = -x + 4.
Further, slope of AC is .
Again, substituting in y = mx + b along with the point ( 1,3 ) gives,
3 = -3 + b i.e. b = 6
So, the equation of AC is y = -3x + 6.
Using 'zero test', we get that,
y = -x + 4 gives 0 = 4, which is not true.
y = -3x + 6 gives 0 = 6, which is not true.
Hence, the solution region will be away from the origin as shown in the figure 1 below.
Thus, we get the system y>-x+4 and y>-3x+6 containing point B and C.
Part B: We have the equations y = -x + 4 and y = -3x + 6.
Substitute the points B = ( 3,1 ) and C = ( 3,-3 ) into these equations respectively, we get,
y = -x + 4 gives y = -3 + 4 i.e. y = 1
y = -3x + 6 gives y = -3 × 3 + 6 i.e. y = -9 + 6 i.e. y = -3.
Hence, points B and C are the solutions of the system in Part A.
Part C: After plotting y>2x+5, we ca see from the second figure that the points included in the shaded region are D = ( -4,2 ) and E = (-1, 5).
So, Lisa is allowed to attend the schools D and E.
