In the adjacent figure $ABC$ is right angled at $B$. If $AB = 4$ and $BC = 3$ and side $AC$ slides along the coordinate axes in such a way that $B$ always remains in the first quadrant, then $B$ always lie on straight line:
Step-by-Step Solution
Key Concept: The complementary angle $\frac{\pi}{2} - \theta$ relates $\tan\theta$ to the perpendicular line's slope via the cotangent function.
Given $\tan\theta = \frac{3}{4}$, the slope of line $OB$ is $\tan(\frac{\pi}{2} - \theta) = \cot\theta$. Since $\tan\theta = \frac{3}{4}$, we have $\cot\theta = \frac{4}{3}$. The locus of point $(h,k)$ on line $OB$ with slope $\frac{4}{3}$ is $k = \frac{4}{3}h$, which simplifies to $3y - 4x = 0$.
<div class="key-concept"><strong>Key Concept:</strong> The complementary angle $\frac{\pi}{2} - \theta$ relates $\tan\theta$ to the perpendicular line's slope via the cotangent function.</div>
<div class="trap-box"><strong>Trap:</strong> Students often confuse the slope of $OA$ with the slope of $OB$; remember that $\cot\theta = \frac{1}{\tan\theta}$, not $\tan\theta$ itself.</div>
Correct Answer: 2