Straight LineHardBloom L3

Question

In the figure shown, $OABC$ is a rectangle with dimensions $OA = 3$ units and $OC = 4$ units. If $AD = 1.5$ units then slope of diagonal $OB$ will be:

Options

A.$\frac{1}{\sqrt{2}}$
B.$\frac{\sqrt{2}}{\sqrt{3}}$
C.$\frac{1}{2}$
D.$\frac{1}{3}$

Solution

Sol. [Key Concept: Use the rectangle's dimensions and the tangent of the diagonal's inclination to get the slope of $OB$.]

$$tan\theta = \frac{3}{4},tan\phi = \frac{3}{2(3)} = \frac{1}{2} $$

$$\begin{matrix} \text{~}\text{slope of}\text{~}OB & \ = \tan\left( \frac{\pi}{2} - (\theta + \phi) \right) = \cot(\theta + \phi) \\ & \ = \frac{1 - \frac{3}{4} \cdot \frac{1}{2}}{\frac{3}{4} + \frac{1}{2}} = \frac{1}{2} \end{matrix}$$

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