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?
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