area of OAB = 72 = (1/2) sin (AOB) * OA * OB solve the above for sin(AOB) to find sin(AOB) = 1/2 area of ODC = 288 = (1/2) sin (DOC) * OD * OD Note that sin(DOC) = sin(AOB) = 1/2, OD = 18 + y and OC = 16 + x and substitute in the above to obtain the first equation in x and y 1152 = (18 + y)(16 + x)...
Y = -6x - 15 We’re already at -3 (y) and the slope is negative so the y intercept is < -3. X is -2 and we need to find the y when x is 0. -2 is 2 away from 0 and 6 (the slope) times 2 is 12. -3 - 12 = - 15
