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Circle O is centered at the origin with point P = (3,4) lying on it.  The red line 3x + 4y - 7 = 0 intersects the circle at A and B.  What is the area of quadrilateral AOBP?

 

 Jan 7, 2020
 #1
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3x +  4y   - 7    =   0        rearrange as

4y  =   -3x + 7

y  =  [ 7  - 3x] / 4   (1)

 

The equation of the  circle is

 

x^2  + y^2  =   25         sub (1)  into this for y

x^2  + [ ( 7  -3x )  / 4 ] ^2  = 25

x^2 +  [ 9x^2  - 42x + 49 ] / 16 =  25        multiply through by 16

16x^2  + 9x^2  - 42x + 49  = 400

25x^2  - 42x - 351   =  0    factor  as

(25x  - 117) ( x + 3)   =  0

Setting the second factor to  0   and solving for    produces  the  x coordinate intersection  of  pont A

So....the x coordinate of  A  =  -3   and the y coordinate of A  =  4

So  A = (-3,4)   P  = (3, 4)  and O =  (0,0)

AP =  sqrt  [ (3  - -3)^2 + ( 4  - 4)^2 ] =  sqrt  [ 6^2 + 0^2]  =  sqrt [ 36 ] =  6

So triangle AOP is isoceles with AP  =6   and  OA, OP   = 5

 

Using Heron's formula, the semi-perimeter   of this triangle  =  [ 5 + 5 + 6] /2=  16 / 2  =  8

 

Andthe area  of this triangle  =

 

sqrt  [8 * (8 - 6) * (8 - 5) * (8 - 5) ] =    sqrt [ 8 * 2 * 3 * 3 ]  =  sqrt [ 144]  =  12  units^2

 

And  using symmetry, [ AOBP ]   is twice this  =   24 units^2

 

 

cool cool cool

 Jan 8, 2020

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