First slide

Equation of line in Straight lines

Question

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

Moderate

A

1/ 4
B

1/ 3
C

1 /2
D

Cannot be determined

Solution

Clearly, $OB = 5,$
So, $\frac{OA}{OB} = \frac{3}{5} = \frac{1 .5}{2 .5} = \frac{AD}{DB}$
Thus, OD bisects $∠ BOC = θ .$ Then $tanθ = \frac{3}{4}$ .
Let $∴ ∠ BOC = \frac{1}{2} ∠ AOB = \frac{1}{2} (90^{0} - θ)$
$∴$ Slope of $OB = \tan \frac{1}{2} (90^{0} - θ) = \frac{\tan 45^{0} - \tan \frac{θ}{2}}{1 + \tan 45^{0} . \tan \frac{θ}{2}}$
Now, $tanθ = \frac{3}{4}$ $\Rightarrow \frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}} = \frac{3}{4}$
$3 - 3 \tan^{2} \frac{θ}{2} = 8 \tan \frac{θ}{2}$
$\Rightarrow 3 \tan^{2} \frac{θ}{2} + 8 \tan \frac{θ}{2} - 3 = 0$
$\Rightarrow (3 \tan \frac{θ}{2} - 1) (\tan \frac{θ}{2} + 3) = 0$
so, $\tan \frac{θ}{2} = \frac{1}{3}$
$∴ Slope of OB = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{1}{2}$

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