Q67.Let R be a rectangle given by the lines ๐ฅ= 0, ๐ฅ= 2, ๐ฆ= 0 and ๐ฆ= 5. Let A๐ผ, 0 and B0, ๐ฝ, ๐ผโ0, 2 and ๐ฝโ0, 5, be such that the line segment ๐ด๐ต divides the area of the rectangle ๐ in the ratio 4: 1. Then, the mid- point of ๐ด๐ต lies on a (1) straight line (2) parabola (3) hyperbola (4) circle
What This Question Tests
The question involves calculating areas of a rectangle and a triangle formed by a line segment, then using the given area ratio to find the locus of the midpoint of that segment.
Concepts Tested
Formulas Used
Area = length * width
Area of triangle = 1/2 * base * height
Midpoint = ((x1+x2)/2, (y1+y2)/2)
๐ NCERT Sections This Tests
2.1 โ Two Charges 5 ร 10โ8 C And โ3 ร 10โ8 C Are Located 16 Cm Apart. At
Physics Class 11 ยท Chapter 2
2.1 Two charges 5 ร 10โ8 C and โ3 ร 10โ8 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 โ Two Charges 2 Mc And โ2 Mc Are Placed At Points A And B 6 Cm
Physics Class 11 ยท Chapter 2
2.3 Two charges 2 mC and โ2 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 โ A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 ยท 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.
๐ Question Details
- Chapter
- Coordinate Geometry
- Topic
- Area, locus, midpoint
- Year
- 2023
- Shift
- 11 Apr Shift 1
- Q Number
- Q67
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 10: Straight Lines
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Q68.A light ray emerging from the point source placed at P(1, 3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R (6, 7), then the abscissa of Q is: (1) 1 (2) 3 (3) 7 (4) 5 2 2
Q76.If two vertices of an equilateral triangle are A(โa, 0) and B(a, 0), a > 0, and the third vertex C lies above x- axis then the equation of the circumcircle of โณABC is : (1) 3x2 + 3y2 โ2โ3ay = 3a2 (2) 3x2 + 3y2 โ2ay = 3a2 (3) x2 + y2 โ2ay = a2 (4) x2 + y2 โโ3ay = a2