in the given figure, OABC is a rectangle. slope of OB is

Clearly, OB=5,So, OAOB=35=1.52.5=ADDBThus, OD bisects ∠BOC=θ. Then tanθ=34.Let ∴∠BOC=12∠AOB=12900−θ∴ Slope of OB=tan12900−θ=tan450−tanθ21+tan450.tanθ2Now, tanθ=34⇒2tanθ21−tan2θ2=343−3tan2θ2=8tanθ2⇒3tan2θ2+8tanθ2−3=0⇒3tanθ2−1tanθ2+3=0so, tanθ2=13∴Slope of OB=1−131+13=12