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

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: $$\cos \theta +\sin \theta =\sqrt{2}\sin (\theta +\frac{\pi }{4})\Rightarrow \sin (\theta +\frac{\pi }{4})>\frac{1}{\sqrt{2}}$$ So: $$\theta \in (0,\frac{\pi }{2})$$

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

Given: Points: $P(1,4)$, $Q(k,3)$

Step 1: Find midpoint of PQ

Midpoint = $({\displaystyle \frac{1+k}{2}},{\displaystyle \frac{4+3}{2}})=({\displaystyle \frac{1+k}{2}},{\displaystyle \frac{7}{2}})$

Step 2: Find slope of PQ

Slope of PQ = ${\displaystyle \frac{3-4}{k-1}}={\displaystyle \frac{-1}{k-1}}$

Step 3: Slope of perpendicular bisector = negative reciprocal = $k-1$

Step 4: Use point-slope form for perpendicular bisector:

$y-{\displaystyle \frac{7}{2}}=(k-1)(x-{\displaystyle \frac{1+k}{2}})$

Step 5: Find y-intercept (put $x=0$)

$y={\displaystyle \frac{7}{2}}+(k-1)(-{\displaystyle \frac{1+k}{2}})$

$y={\displaystyle \frac{7}{2}}-(k-1)\left({\displaystyle \frac{1+k}{2}}\right)$

Given: y-intercept = -4, so:

${\displaystyle \frac{7}{2}}-{\displaystyle \frac{(k-1)(k+1)}{2}}=-4$

Multiply both sides by 2:

$7-(k^2-1)=-8\Rightarrow 7-k^2+1=-8\Rightarrow 8-k^2=-8$

$\Rightarrow k^2=16\Rightarrow k=\pm 4$

✅ Final Answer: $\boxed{{\displaystyle k=-4\text{ or }4}}$