NIMCET Straight Line Previous Year Questions (PYQs) – Page 1 of 3

Given: Lines   $4ax+2ay+c=0$   and   $5bx+2by+d=0$.

Condition: Intersection lies in the 4th quadrant and is equidistant from the axes $\Rightarrow$ point is of the form $(t,-t)$ with $t>0$.

Substitute $y=-x$ in the two lines:
$4ax+2a(-x)+c=0 \;\Rightarrow\; 2x+\dfrac{c}{2a}=0 \;\Rightarrow\; x=-\dfrac{c}{2a}.$
$5bx+2b(-x)+d=0 \;\Rightarrow\; 3x+\dfrac{d}{b}=0 \;\Rightarrow\; x=-\dfrac{d}{3b}.$

Equate $x$ from both:   $-\dfrac{c}{2a}=-\dfrac{d}{3b} \;\Rightarrow\; \boxed{3bc=2ad}.$

(QIV check: $x>0,\ y<0 \;\Rightarrow\; \dfrac{c}{a}<0$ and $\dfrac{d}{b}<0$ which is consistent.)

Same Side of a Line — Geometric Condition

Line: \( x + y - 1 = 0 \)
Point 1: (3, 2) → Lies on the side where value is positive:
\[ f(3, 2) = 3 + 2 - 1 = 4 > 0 \]

Point 2: \( (\cos\theta, \sin\theta) \) lies on same side if: \[ \cos\theta + \sin\theta > 1 \] Using identity: \[ \cos\theta + \sin\theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \Rightarrow \sin\left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}} \] So: \[ \theta \in \left(0,\ \frac{\pi}{2}\right) \]

✅ Final Answer: \( \boxed{\theta \in \left(0,\ \frac{\pi}{2}\right)} \)

This is a diamond (square rotated 45°) with diagonals length $2$ and $2$. Area $=\dfrac12 d_1 d_2=\dfrac12\cdot2\cdot2=2$

Intersection point $(x,y)$ satisfies: 

Axis intercepts of both lines lie on same circle. Correct center is $ \left(\dfrac52,\dfrac52\right) $.