JEE Maths›Straight Lines›Visual Solution
Visual SolutionPYQ 2024 · Jan Shift 2Tricky
Question
Let R be the interior region between the lines 3x−y+1=03x - y + 1 = 0 and x+2y−5=0x + 2y - 5 = 0 containing the origin. The set of all values of aa, for which the points (a2,a+1)(a^2, a+1) lie in RR, is:
(A)(−3,−1)∪(−13,1)(-3,-1) \cup \left(-\frac{1}{3}, 1\right)
(B)(−3,0)∪(13,1)(-3,0) \cup \left(\frac{1}{3}, 1\right)
(C)(−3,0)∪(23,1)(-3,0) \cup \left(\frac{2}{3}, 1\right)
(D)(−3,−1)∪(13,1)(-3,-1) \cup \left(\frac{1}{3}, 1\right)
Solution Path
Same side conditions: L1L_1 gives a(3a−1)>0a(3a-1)>0, L2L_2 gives (a+3)(a−1)<0(a+3)(a-1)<0. Intersection: (−3,0)∪(1/3,1)(-3,0) \cup (1/3, 1).
01Question Setup
1/4
Find the values of aa for which the origin lies on the same side of two given lines.
a∈  ?a \in \;?
02Same Side as Origin
2/4
L1L_1: 3a2−a>03a^2 - a > 0 gives a∈(−∞,0)∪(13,∞)a \in (-\infty, 0) \cup (\tfrac{1}{3}, \infty). L2L_2: a2+2a−3<0a^2 + 2a - 3 < 0 gives a∈(−3,1)a \in (-3, 1).
L1:a(3a−1)>0L_1: a(3a-1) > 0, L2:(a+3)(a−1)<0L_2: (a+3)(a-1) < 0
03IntersectionKEY INSIGHT
3/4
Intersect the two sets: (−3,0)∪(13,1)(-3, 0) \cup (\tfrac{1}{3}, 1).
a∈(−3,0)∪(13,1)a \in (-3, 0) \cup \left(\frac{1}{3}, 1\right)
04Final Answer
4/4
The intersection of both conditions gives the final answer.
(−3,0)∪(13,1)\boxed{(-3, 0) \cup \left(\tfrac{1}{3}, 1\right)}
Concepts from this question1 concepts unlocked
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Same Side of a Line Test

EASY

Two points are on the same side of ax + by + c = 0 if they give the same sign when substituted.

(ax1+by1+c)(ax2+by2+c)>0(ax_1 + by_1 + c)(ax_2 + by_2 + c) > 0

Key for region problems where you need to check which side of a line a point lies on.

Region problemsInequality constraintsLocus
Practice (2 Qs) →