Q65.Let A(−1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of △PAB is 10 . If the locus of P is ax + by = 15, then 5a + 2 b is : (1) 6 (2) −65 (3) 4 (4) −125
What This Question Tests
This problem requires calculating the area of a triangle using the base-height formula, where the height is the perpendicular distance from the variable point P to the line AB, to find the locus of P.
Concepts Tested
Formulas Used
Area = (1/2) * base * height
Distance from point (x₀, y₀) to Ax+By+C=0 is |Ax₀+By₀+C|/√(A²+B²)
📚 NCERT Sections This Tests
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
2.3 — Two Charges 2 Mc And –2 Mc Are Placed At Points A And B 6 Cm
Physics Class 11 · Chapter 2
2.3 Two charges 2 mC and –2 mC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?
2.2 — A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 · Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
📋 Question Details
- Chapter
- Straight Lines
- Topic
- Area of a triangle
- Year
- 2024
- Shift
- 05 Apr Shift 2
- Q Number
- Q65
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 10: Straight Lines
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