Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade 11
Question:
A line '$L$' is drawn from $(4, 3)$ to meet the lines $L_1: 3x + 4y + 5 = 0$ and $L_2: 3x + 4y + 15 = 0$ at points $A$ and $B$ respectively. From point '$A$', a line perpendicular to $L$ is drawn meeting the line '$L_2$' at $A_1$. Similarly, from point '$B$' a line perpendicular to $L$ is drawn meeting the line $L_1$ at $B_1$. Thus a parallelogram $AA_1B_1B$ is formed. If the area of the parallelogram $AA_1B_1$ is least, the equation of the line $L$ is/are:
$x - 7y + 17 = 0$
$x + 7y + 1 = 0$
$3x + y - 31 = 0$
$7x + 2y - 31 = 0$
Step-by-Step Solution
Key Concept: Coordinate transformation simplifies constraints; eliminating variables reveals geometric bounds on triangle dimensions.
The area of parallelogram $AA_1B_1B$ is given by $\frac{8}{\sin 2\theta}$. For $\theta = \frac{\pi}{4}$, we have $\sin 2\theta = 1$, so area $= 8$. The constraint $b + c > a$ combined with $\cos B = \frac{b^2+c^2-36}{2bc} 6$, giving $b > 2$ and $2 a$ gives $b + 2h > 6$, so $b > 2$. From $\cos B = \frac{b^2+c^2-36}{2bc} < 1$ with $c = 2b$, we get $b < 6$. Using $h^2 + k^2 = c^2$ and $(h-6)^2 + k^2 = b^2$, eliminating $b^2$ yields $k^2 = 16-(h-8)^2$, giving $\max k = 4$.
Correct Answer: 1,3