Question

*******ASAP******

Answer 1

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 $\frac{1-3}{3-1} =\frac{-2}{2} =-1$.

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 $\frac{-3-3}{3-1} =\frac{-6}{2} =-3$.

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.