Q66.Let a triangle be bounded by the lines L1 : 2x + 5y = 10 ; L2 : โ4x + 3y = 12 and the line L3 , which passes through the point P(2, 3), intersect L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to (1) 110 (2) 132 13 13 (3) 142 (4) 151 13 13
What This Question Tests
The problem requires finding the vertices of a triangle by using line equations and the section formula, then calculating the triangle's area.
Concepts Tested
Formulas Used
Section formula: ((m x2 + n x1) / (m+n), (m y2 + n y1) / (m+n))
Area of triangle = 0.5 * |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
Equation of line passing through two points
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.
3.10 โ In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 ยท Chapter 3
3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol Lโ1 0.20 0.20 0.40 B/ mol Lโ1 0.30 0.10 0.05 r0/mol Lโ1sโ1 5.07 ร 10โ5 5.07 ร 10โ5 1.43 ร 10โ4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ยฎ C + D Experiment [A]/mol Lโ1 [B]/mol Lโ1 Initial rate of formation of D/mol Lโ1 minโ1 I 0.1 0.1 6.0 ร 10โ3 II 0.3 0.2 7.2 ร 10โ2 III 0.3 0.4 2.88 ร 10โ1 IV 0.4 0.1 2.40 ร 10โ2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol Lโ1 [B]/ mol Lโ1 Initial rate/ mol Lโ1 minโ1 I 0.1 0.1 2.0 ร 10โ2 II โ 0.2 4.0 ร 10โ2 III 0.4 0.4 โ IV โ 0.2 2.0 ร 10โ2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 sโ1 (ii) 2 minโ1 (iii) 4 yearsโ1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ยฎ 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 ร [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol Lโ1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
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 of a triangle
- Year
- 2022
- Shift
- 28 Jun Shift 2
- Q Number
- Q66
- 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) โ2 (4) โ4
Q72.The normal at (2, 23 ) to the ellipse, x216 + y23 = 1 touches a parabola, whose equation is (1) y2 = โ104x (2) y2 = 14x (3) y2 = 26x (4) y2 = โ14x sin(ฯ 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(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