Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade None
Question:
Let $B(1,-3)$ and $D(0,4)$ represent two vertices of rhombus $ABCD$ in $(x, y)$ plane, then coordinates of vertex $A$ is $\angle BAD = 60°$ can be equal to:
\left(\frac{1-7\sqrt{3}}{2}, \frac{1-\sqrt{3}}{2}\right)
\left(\frac{1+7\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)
\left(\frac{-1+7\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right)
\left(\frac{-1-7\sqrt{3}}{2}, \frac{-1-\sqrt{3}}{2}\right)
Step-by-Step Solution
Key Concept: Use trigonometric values to find point coordinates through rotation and distance formulas.
Given $\tan\theta = \frac{1}{7}$, $\sin\theta = \frac{1}{5\sqrt{2}}$, $\cos\theta = \frac{7}{5\sqrt{2}}$, we find $AP = \frac{7}{2}\sqrt{2}\cot 30° = \frac{7}{2}\sqrt{2} \cdot \sqrt{3} = \frac{7\sqrt{6}}{2}$. The coordinates of point $A$ are calculated using the angle of rotation, yielding two possible positions: $A = \left(\frac{1-7\sqrt{3}}{2}, \frac{1-\sqrt{3}}{2}\right)$ or $A = \left(\frac{1+7\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)$.
Correct Answer: 1,2