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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
If (x- y) `in Q_(4)` , then (x , y) belongs to which quadrant ? |
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Answer» Given, `(x , -y) in Q_(4) implies x gt 0 , y lt 0` . `therefore ` (x , y) belongs to first quadrant , i.e., `Q_(1)`. |
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| 52. |
If the coordinates of the two points are P(-2,3) and Q(-3,5) , then (Abscissa of P)-(Abscissa of Q) isA. `-5`B. 1C. `-1`D. `-2` |
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Answer» Correct Answer - B We have, points P(-2,3) and Q(-3,5) Here, abscissa of P i.e., x-coordinate of P is -2 and abscissa of Q i.e., x-coordinate of Q is -3 So, (Abscissa of P)-(Abscissa of Q)=-2-(-3)=-2+3=1 |
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| 53. |
If `A(-2, 3) and B(-3, 5)` are two given points then (abscissa of A) - (abscissa of B) = ?A. `-2`B. 1C. `-1`D. 2 |
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Answer» Correct Answer - B (abscissa of A) - (abscissa of B) `= (-2) -(-2 + 3) = 1` |
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| 54. |
The area of a triangle with vertices A(3,0),B(7,0) and C(8,4) isA. 14B. 28C. 8D. 6 |
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Answer» Correct Answer - c Area of `triangleABC` whose Vertices A `-=(x_(1),y_(1)),B-=(x_(2),y_(2))and C-=(x_(3),y_(3))` are given by `triangle` `=|(1)/(2)[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))]|` Here , `x_(1)=3,y_(1)=0,x_(2)=7,y_(2)=0x_(3)=8 and y_(3)=4` `:.` triangle=|(1)/(2)[3(0-4)+7(4-0)+8(0-0)]|=| (1)/(2)(-12+28+0)|(1)/(2)(16)|=8` Hence , the requirred area of `triangleABC` is 8. |
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| 55. |
If slope of a line (l) is `tan theta` , then slope of a line perpendicular to (l) is `"_______"` |
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Answer» Correct Answer - `-cot theta ` |
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| 56. |
The lines x = 2 and y = - 3 intersect in `"_______"` quadrant . |
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Answer» Correct Answer - fourth |
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| 57. |
The angle made by a line with the positive direction of X-axis is `45^(@)` . Complete the following activity to find the slope of the line . |
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Answer» Angle made by the line with the positive direction of X-axis ` (theta) = square ` Slope of the line `square ` = `tan square ` ` = square " " ` ( Writing the value ) Slope of the line is `square ` Angle make by the line with the positive direction of X-axis Slope of the line = `(theta)=45^(@)` Slope of the line `= tan theta ` `= tan 45^(@)` `=1 ` ... (Writing the value ) Slope of this line is 1 . |
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| 58. |
A line makes an anlge of `30^(@)` with the positive direction of X-axis. So the slope of the line is _ _ _ _ _ _ _ _A. `(1)/(2)`B. `(sqrt(3))/(2)`C. `(1)/(sqrt(3))`D. `sqrt(3)` |
| Answer» Correct Answer - C | |
| 59. |
Abscissa of a point is positive inA. I and II quadrantsB. I and IV quadrantsC. I quadrant onlyD. II quadrant only |
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Answer» Correct Answer - B Abscissa of a point is positive in I and IV quadrants |
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| 60. |
The points `(-5, 3) and (3, -5)` lie in theA. same quadrantB. II and III quadrants respectivelyC. II and IV quadrants respectivelyD. IV and II quadrants respectively |
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Answer» Correct Answer - C `(-5, 3) and (3, -5)` lie in II and IV quadrants respectively. |
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| 61. |
Find the slope of the line joining points (3,8) and `(-9 , 6)` . |
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Answer» Let A( 3, 8) and B(-9 , 6) be the given points . Then the slope of `bar (AB) = (y_(2) - y_(1))/(x_(2) - x_(1))` `= (6-8)/(-9-3) = (1)/(6)`. |
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| 62. |
The points (-4,0),(4,0) and (0,3) are the verticess of aA. right angled triangleB. isosceles triangleC. equilateral triangleD. scealene triangle |
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Answer» Correct Answer - b Let A(-4,0),B(4,0),C(0,3) are the given vertices. Now , distance between A(-4,0) and B(4,0), `AB=sqrt([4-(-4)]^(2)+(0-0)^(2))` [`:.` distance between two points `(x_(1),y_(1) and (x_(2),y_(2)) d=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]` `=sqrt((4+4)^(2))=sqrt(8^(2))=8` Distance between B(4,0) and C(0,3), and BC `=sqrt((0-4)^(2)+(3-0)^(2))=sqrt(16+9)=sqrt(25)=5` Distance between A(-4,0) and C(0,3), AC `=sqrt([0-(-4)]^(2)+(3-0)^(2))=sqrt(16+9)=sqrt(25)=5` `:.` BC =AC Hence , `triangleABC` is an isosceles triangle because an isosceles triangle has two sides equal. |
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| 63. |
If A (1,3),B(-1,2),C(2,5) and D(x,4) the vertices of `square ABCD ` them find the value of x . |
| Answer» Correct Answer - A | |
| 64. |
A(4,7) and B (2,1), P(3,a) is the midpoint of seg AB, then the value of a is _ _ _ _ _ _A. 4B. 8C. 6D. 3 |
| Answer» Correct Answer - A | |
| 65. |
If the slope of a line is `sqrt(3) `, the angle made by the line with the positive direction of X-axis is _ _ _ _ _ _A. `60^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)` |
| Answer» Correct Answer - A | |
| 66. |
The point at which the two coordinate axes meet is called theA. abscissaB. ordinateC. originD. quadrant |
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Answer» Correct Answer - C The two coordinate axes meet at the origin |
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| 67. |
Which of the following points lies on the line `y = 2x + 3` ?A. (2, 8)B. (3, 9)C. (4, 12)D. (5, 15) |
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Answer» Correct Answer - B Clearly, (3,9) satisfies `y = 2x + 3` |
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| 68. |
Points `(1, -1), (2, -2), (-3, -4), (4, -5)`A. all lie in the II quadrantB. all lie in the III qudrantC. all lie in the IV quadrantD. do not lie in the same quadrant |
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Answer» Correct Answer - D `(1, -1), (2, -2), (-3, -4), (4, -5)` do not lie in the same quadrant |
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| 69. |
Which of the following points does not lie on the line `y = 3x + 4` ?A. (1, 7)B. (2, 10)C. `(-1, 1)`D. (4, 12) |
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Answer» Correct Answer - D Clearly, (4, 12) does not satisfy `y = 3x + 4` |
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| 70. |
The point which divides the line segment joining the points (7,-6) and (3,4) in ration ` 1 : 2` internally lies in theA. I quadrantB. II quadrantC. III quadrantD. Iv quadrant |
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Answer» Correct Answer - d If P(x,y) divides the line segment joining A`(x_(1),y_(1))and B(x_(2),y_(2))` internally in the ration m: n, then `x=(mx+nx_(1))/(m+n) and y=(my_(2)+ny_(1))/(m+n)` Given that , `x_(1)=7,y_(1)=-6,x_(2)=3,y_(2)=4,m=1and n=2` `:.` `x=(1(3)+2(7))/(1+2),y=(1(4)+2(-6))/(1+2)` [by section formula] `rArr x=(3+14)/(3),y=(4-12)/(3)` `rArr(17)/(3),y=-(8)/(3)` So , (x,y) `=(17/(3),-(8)/(3))` lines in IV quadrant. [Since , in quadrant , x-coordinate is positive and y-coordinate is negative] |
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| 71. |
The slope of line joining point P(-1,1) and Q (5,-7) is _ _ _ _ _ _ _ _A. `-4/3`B. 16C. -3D. -2 |
| Answer» Correct Answer - A | |
| 72. |
Points `(1, -1), (2, -2), (-3, -4), (4, -5)`A. lie in II quadrantB. lie in III quadrantC. lie in IV quadrantD. do not lie in the same quadrant |
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Answer» Correct Answer - D In points (1,-1), (2,-2) and (4,-5) x-coordinate is positive and y-coordinate is positive and y-coordinate negative , So, they all lie in IV quadrant . In point(-3,-4) x-coordinate is negative and So, given points do not lie in the same quadrant. |
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| 73. |
The point whose ordinate is 3 and which lies on the y-axis isA. (3,0)B. (0, 3)C. (3, 3)D. (1, 3) |
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Answer» Correct Answer - B A point lying on the y-axis with ordinate 3 is (0, 3) |
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| 74. |
Point `(0, -8)` liesA. in the II quadrantB. in the IV quadrantC. on the x-axisD. on the y-axis |
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Answer» Correct Answer - D `(0, -8)` lies on the y-axis |
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| 75. |
The point whose ordinate is 4 and which lies on y-axis isA. (4,0)B. (0,4)C. (1,4)D. (4,2) |
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Answer» Correct Answer - B Given ordinate of the point is 34 and the point lies on Y-axis , so its abscissa is zero. Hence , the required point is (0,4). |
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| 76. |
In which quadrant or on which axis do each of the points `(2, 4), (3, 1), (1, 0),(1,2) a n d (3, 5)`lie? Verify your answer by locating them on the Cartesian plane. |
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Answer» In first quadrant, value of `x` coordinate is positive and value of `y` coordinate is positive. In second quadrant, value of `x` coordinate is negative and value of `y` coordinate is positive. In third quadrant, value of `x` coordinate is negative and value of `y` coordinate is negative. In fourth quadrant, value of `x` coordinate is positive and value of `y` coordinate is negative. On the basis of above, Point(-2,4) lies in second quadrant. Point(3,-1) lies in fourth quadrant. Point(-1,0) lies on x-axis as y coordinate is `0` Point(1,2) lies in first quadrant. Point(-3,-5) lies in third quadrant. |
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| 77. |
Which of the following points does not lie in any quadrant?A. `(3, -6)`B. `(-3, 4)`C. (5, 7)D. (0, 3) |
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Answer» Correct Answer - D The point (0, 3) does not lie in any quadrant |
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| 78. |
The point which lies on the perpendicular bisector of the line segment joining the points A(-2,-5) and B (2,5) isA. (0,0)B. (0,2)C. (2,0)D. (-2,0) |
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Answer» Correct Answer - a We know that , the perpendicular bisector of the any line segment divides the line segment into two equal parts i.e., the perpendicular bisector of the line segment always passes through the segment joining the points A(-2,-5) and B(2,5) `=((-2+2)/(2),(-5+5)/(2))=(0,0)` Hence , (0,0) is the required point lies on the perpendicular bisector of the lines segment . |
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| 79. |
A school bus driver starts from the school drives 2 km towards North, takes a left turn and drives for 5 km. He then takes a left turn and drives for 8 km before taking a left turn again and driving for further 5 km. The driver finally takes a left turn and drives 1 km before stopping. How far and towards which direction should the driver drive to reach the school again? |
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Answer» school south direction `5km` away `5 km`north side walk Answer |
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| 80. |
The point (-5,2) and (2,-5) lie in theA. same quadrantB. II and III quadrants , respectivelyC. II and IV quadrants , respectivelyD. IV and II quadrants, respectively |
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Answer» Correct Answer - C In point (-5,2) , x-coordinate is negative and y-coordinate is positive , so it lies in II quadrant and in point (2,-5) , x-coordinate is positive and y-coordinate is negative so it lies in IV quadrant |
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| 81. |
Point `(-7, 0)` liesA. on the negative direction of the x-axisB. on the negative direction of the y-axisC. in the III quadrantD. in the IV qudrant |
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Answer» Correct Answer - A `(-7, 0)` lies on the negative direction of the x-axis |
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| 82. |
The ordinate of every point on the x-axis isA. 1B. `-1`C. 0D. any real number |
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Answer» Correct Answer - C Every point on the x-axis is of the form (x, 0) and therefore its ordinate is 0 |
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| 83. |
A point both of whose coordinates are negative lies in quadrantA. IB. IIC. IIID. IV |
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Answer» Correct Answer - C A point both of whose coordinates are negative lies in III quadrant |
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| 84. |
If (x,y) represents a point and `|x| gt 0` and `y lt 0` , then in which quadrant(s) can the point lie ? |
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Answer» Correct Answer - The point may lie in `Q_(3)` or `Q_(4)`. |
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| 85. |
If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis then the point P hasA. x-coordinate =-5B. y-coordinate =5 onlyC. y-coordiante=-5onlyD. y-coordinate =5 or -5 |
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Answer» Correct Answer - D We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be meausre in II quadrant or III quadrant . Hence , the point P has y-coordinate =5 or -5. |
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| 86. |
The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis isA. `(-5, 0)`B. `(0, -5)`C. `(5, 0)`D. `(0, 5)` |
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Answer» Correct Answer - B The required point is `(0, -5)` |
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| 87. |
Write the axis on which the given point lies. (i) (2, 0) (ii) `(0, -5)` (iii) `(-4, 0)` (d) (0, -1) |
| Answer» Correct Answer - (i) x-axis (ii) y-axis (iii) x-axis (iv) y-axis | |
| 88. |
For each of the following points write the quadrant in which it lies. (i) `(-6, 3)` (ii) `(-5, -3)` (iii) (11, 6) (iv) `(1, -4)` (v) `(-7, -4)` (vi) `(4, -1)` (vii) `(-3, 8)` (viii) `(3, -8)` |
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Answer» Correct Answer - (i) II (ii) III (iii) I (iv) IV (v) III (vi) IV (vii) II (viii) IV |
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| 89. |
If `x gt 0 and y lt 0` then the point (x, y) lies in quadrantA. IB. IIIC. IID. IV |
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Answer» Correct Answer - D If `x gt 0 and y ly 0` then the point (x, y) lies in IV quadrant |
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| 90. |
The two lines 3x + 4y - 6 = 0 and 6x + ky - 7 = 0 are such that any line which is perpendicular to the first line is also perpendicular to the second line . Then , k = `"_______"`.A. `-2`B. `-6`C. 6D. 8 |
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Answer» Correct Answer - D If two lines are perpendicular to the same line , then they are parallel to each other . |
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| 91. |
Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and(-k, -5) is 24 sq. units. |
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Answer» Let A(1, -1), B(-4, 2k) and C(-k, -5) be the vertices of the given `Delta ABC.` Then, `A(x_(1) = 1, y_(1) = -1), B(x_(2) =-4, y_(2) = 2k), " and " C(x_(3) = -k, y_(3) = -5)`. `therefore ar(Delta ABC) = (1)/(2) |x_(1) (y_(2) - y_(3)) + x_(2) (y_(3)-y_(1)) +x_(3)(y_(1) -y_(2))|` `= (1)/(2)|*(2k+5) -4 * (-5+1) -k(-1) -2k)|` `=(1)/(2)|2k + 5 +16 +k + 2k^(2)|` `= (1)/(2)|2k^(2) + 3k +21|` sq units. But, it is given that `ar(Delta ABC)` = 24 sq units. `therefore (1)/(2)|2k^(2) + 3k + 21| =24 rArr |k^(2) + (3)/(2) k + (21)/(2)| = 24.` But `{k^(2) + (3)/(2)k + (21)/(2)} = (k^(2) + (3)/(2)k + (9)/(16)) + ((21)/(2) - (9)/(16)) = {(k + (3)/(2))^(2) + (159)/(16)} gt 0.` So, we may write (i) as `k^(2) + (3)/(2)k + (21)/(2) = 24 rArr 2k^(2) + 3k+ 21 = 48` `rArr 2k^(2) +3k -27 = 0 rArr 2k^(2) +9k-6k -27 = 0` `rArr k(2k+9)-3(2k +9) = 0rArr (2k +9)(k-3) = 0` `rArr k -3 = 0 "or" 2k +9 =9 rArr k = 3 "or" k = -(9)/(2)` Hence, k = 3 or`k = -(9)/(2).` |
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| 92. |
The radius of a circle with centre (-2,3 ) is 5 units , then the point (2,5) lies `"_______"`.A. on the circleB. inside the circleC. outside the circleD. None of the above. |
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Answer» Correct Answer - B Find the distance between the given two points and compare that distance with the radius given. |
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| 93. |
The line x = my , where m `lt` 0 , lies in the quadrants .A. 1st , 2ndB. 2nd , 4thC. 3rd , 4thD. 3rd , 1st |
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Answer» Correct Answer - B Identify the sign of y for each sign of x . |
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| 94. |
If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right triangle with `angleB = 90^(@)` then find the value of t. |
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Answer» Correct Answer - t = 1 `AB^(2)+BC^(2) = AC^(2) rArr 25 + {16 + (t + 2)^(2)} = 49 + (t +2)^(2)` `rArr (t+2)^(2) -(t-2)^(2) = 8 rArr 4 xx t xx 2 = 8 rArr t = 1.` |
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| 95. |
Show that points A(3,-1) , B(-1 , 2) and C (6,3) form an isosceles right -angled triangle when joined . |
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Answer» Given , A = (3 , -1) , B(-1, 2) and C = (6,3). AB = `sqrt((-1-3)^(2) + (2+1)^(2)) = 5` units BC = `sqrt((6-(-1))^(2) + (3-2)^(2)) = sqrt(50) `units AC = `sqrt((6-3)^(2) + (3-(-1))^(2)) = 5 ` units Clearly , `BC^(2) = AB^(2) + AC^(2)` . Also , AB = AC. Hence , the given points from the vertices of a right-angled isosceles triangle . |
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| 96. |
If the points A(4, 3) and B(x, 5) lie on a circle with the centre O(2, 3), find the value of x. |
| Answer» Correct Answer - x=2 | |
| 97. |
Find the area in square units , of the rhombus with vertices (2,1) , (-5,2) , (-4, -5) and (3,-6) , taken in that order .A. 24B. 48C. 36D. 50 |
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Answer» Correct Answer - B Find lengths of the diagonals , then area of rhombus = `(1)/(2) xx d_(1) xx d_(2)` . |
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| 98. |
Show that the points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus. |
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Answer» Correct Answer - 45 sq units Diagonal `AC = sqrt((-5)^(2) +5^(2)) = sqrt(50) = 5sqrt(2)`, and diagonal `BD = sqrt(9^(2) + 9^(2)) = sqrt(162) = 9sqrt(2).` `therefore " area " = ((1)/(2) xx AC xx BD) = ((1)/(2) xx 5sqrt(2) xx 9sqrt(2))` sq units = 45 sq units. (Hint : Area of a rhombus `= (1)/(2) xx` (product of its diagonals). |
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| 99. |
Show that the points A(-5, 6), B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area. |
| Answer» Correct Answer - 50 sq units | |
| 100. |
Show that the points A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) are the vertices of a rhombus. Find its area. |
| Answer» Correct Answer - 24 sq units | |