JEE Maths›Straight Lines›Visual Solution
Visual SolutionPYQ 2024 · Jan Shift 2Standard
Question
Let A and B be two finite sets with m and n elements respectively. The total number of subsets of A is 56 more than the total number of subsets of B. Then the distance of the point P(m,n)P(m, n) from the point Q(−2,−3)Q(-2, -3) is:
(A)1010
(B)66
(C)44
(D)88
Solution Path
2n(2m−n−1)=23×72^n(2^{m-n} - 1) = 2^3 \times 7, so n=3n=3, m=6m=6. P(6,3)P(6,3) to Q(−2,−3)Q(-2,-3): distance =64+36=10= \sqrt{64+36} = 10.
01Question Setup
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Set AA has mm elements, set BB has nn elements. 2m−2n=562^m - 2^n = 56. Find distance from P(m,n)P(m,n) to Q(−2,−3)Q(-2,-3).
Find PQPQ given 2m−2n=562^m - 2^n = 56
02Find m, n
2/4
Factor: 2n(2m−n−1)=56=23×72^n(2^{m-n} - 1) = 56 = 2^3 \times 7. So n=3n = 3 and 2m−3=82^{m-3} = 8, giving m=6m = 6.
m=6,  n=3m = 6, \; n = 3
03Distance FormulaKEY INSIGHT
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P(6,3)P(6,3), Q(−2,−3)Q(-2,-3). PQ=(6−(−2))2+(3−(−3))2=64+36=100=10PQ = \sqrt{(6-(-2))^2 + (3-(-3))^2} = \sqrt{64 + 36} = \sqrt{100} = 10.
PQ=82+62=100=10PQ = \sqrt{8^2 + 6^2} = \sqrt{100} = 10
04Final Answer
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Two-part problem: number theory to find coordinates, then distance formula.
PQ=10PQ = \boxed{10} - Answer (A)
Concepts from this question1 concepts unlocked
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Distance from Point to Line

EASY

Distance from (x₁, y₁) to line ax + by + c = 0 is |ax₁ + by₁ + c| / √(a² + b²).

d=∣ax1+by1+c∣a2+b2d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Used in area calculations, mirror image, and foot of perpendicular problems. High-frequency JEE formula.

Distance problemsArea of triangleFoot of perpendicular
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