Q77.Let the orthocentre and centroid of a triangle be A(−3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is: (1) 3√5 (2) √10 2 (3) 2√10 (4) 3√52
What This Question Tests
This question tests the knowledge of the Euler line property, stating that the centroid divides the line segment joining the orthocentre and circumcentre in a 2:1 ratio. It then requires applying the distance formula to find the radius of a circle.
Concepts Tested
Formulas Used
G = (H + 2O)/3 (Section formula for Euler line)
Distance = √((x₂-x₁)² + (y₂-y₁)²)
📚 NCERT Sections This Tests
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.
12.7 — The Radius Of The Innermost Electron Orbit Of A Hydrogen Atom Is
Physics Class 12 · Chapter 12
12.7 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?
5.11 — Draw All The Isomers (Geometrical And Optical) Of:
Chemistry Class 11 · Chapter 5
5.11 Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2] + (ii) [Co(NH3)Cl(en)2] 2+ (iii) [Co(NH3)2Cl2(en)]+
📋 Question Details
- Chapter
- Coordinate Geometry
- Topic
- Properties of triangle centers
- Year
- 2018
- Shift
- 08 Apr
- Q Number
- Q77
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 10: Straight Lines
More from this Chapter
Q78.The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept - 4. Then a possible value of k is (1) 1 (2) 2 (3) −2 (4) −4
Q72.The normal at (2, 23 ) to the ellipse, x216 + y23 = 1 touches a parabola, whose equation is (1) y2 = −104x (2) y2 = 14x (3) y2 = 26x (4) y2 = −14x sin(π cos2 x)
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(−a, 0) and B(a, 0), a > 0, and the third vertex C lies above x- axis then the equation of the circumcircle of △ABC is : (1) 3x2 + 3y2 −2√3ay = 3a2 (2) 3x2 + 3y2 −2ay = 3a2 (3) x2 + y2 −2ay = a2 (4) x2 + y2 −√3ay = a2