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(a) The line x + y = 3 intersects the parabola y = x^2 + 1 at points A and B.  Determine the coordinates of A and B.

 

(b) If C is the vertex of the given parabola, determine the area of triangle ABC.

 Dec 5, 2019
 #1
avatar+109064 
+1

x +  y  =  3   ⇒  y =  3 - x      (1)

y  = x^2 + 1    (2)

 

Equate (1)  and (2)  and we have that

 

3 - x  = x^2 + 1   rearrange as

 

x^2 + x - 2   =  0          factor

 

(x + 2) ( x - 1)  =  0   Set each factor to 0 and solve for  x  and we have that

 

x = -2     and x  =  1

 

So  when x  = -2,  x + y  = 3  implies that y = 5

And when x = 1,  x + y  = 3  implies that y  = 2

 

So   A  = ( -2, 5)   and B  = ( 1, 2)

 

 

cool cool  cool

 Dec 5, 2019
 #2
avatar+109064 
+1

Second one

 

The vertex  of the parabola  is  (0, 1)

 

So   A  = (-2, 5)     B  = (1, 2)     C  = ( 0,1)

 

We can let AB  be the base of the triangle

 AB  =  sqrt [ ( -2 - 1)^2 + (5 - 2)^2 ]  =  sqrt [ 3^2  + 3^2]  = sqrt  [18]  =  3sqrt (2)

 

 

And   we can use the formula for the distance that  (0, 1)  is from the line  x + y - 3 =  0.....this will be the height of ABC

 

So we have

 

l Ax + By - C  l

_____________    =

sqrt ( A^2 + B^2)

 

l   1(0)  + 1(1)   -  3   l                  l  -2  l                2

_________________  =           _______  =       ______  =     sqrt  (2)

   sqrt  [ 1^2 + 1^2]                      sqrt(2)             sqrt (2)

 

 

So.....the area of  ABC  =    (1/2) AB * height of ABC  =  (1/2)(3 sqrt (2) ) ( sqrt (2) )  =  (1/2) (3) (2)  =  

 

 

3 units^2  =  [ ABC  ]

 

 

cool cool cool

 Dec 5, 2019

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