In the given figure, OABC is a rectangle. Slope of OB is
a. `1//4` b. `1//3` c. `1//2` d. Cannot be determined

Options

A. `1//4`
B. `1//3`
C. `1//2`
D. Cannot be determined

Answer

C

Explanation

Clearly, `OB=5`
So, `(OA)/(OB)=(3)/(5)=(1.5)/(2.5)=(AD)/(DB)`
Thus, OD bisects `angle AOB`.
Let `angleBOC=theta`. Then `tantheta=(3)/(4)`
`therefore angleBOC=(1)/(2)angleAOB=(1)/(2)(90^@-theta)`
`therefore` Slope of `OB=tan (1)/(2)(90^@-theta)=(tan45^@-tan.(theta)/(2))/(1+tan45^@+tan.(theta)/(2))`
Now, `tantheta=(3)/(4)`
`rArr (2tan.(theta)/(2))/(1-tan^2.(theta)/(2))=(3)/(4)`
`3-3tan^2.(theta)/(2)=8tan.(theta)/(2)`
`rArr3tan^2.(theta)/(2)+8tan.(theta)/(2)-3=0`
`rArr(3tan.(theta)/(2)-1)(tan.(theta)/(2)+3)=0`
So, `tan.(theta)/(2)=(1)/(3)" "(because(theta)/(2) "is acute")`
`therefore` Slope of `OB =(1-(1)/(3))/(1+(1)/(3))=(1)/(2)`