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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.
| 1. |
Find the equation of a line , whose y-intercept is `-5` and passes through point A (-3,2) . |
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Answer» Correct Answer - 7x + 3y + 15 = 0 |
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| 2. |
Find the equation of the line that passes through point `(5,-3)` and makes an intercept 4 on the X-axis .A. `3x - y + 12 = 0`B. `3x + y + 12 = 0`C. `3x - y -12 = 0`D. `3x + y - 12 = 0` |
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Answer» Correct Answer - D If x-intercept is a ,then (a ,0) is a point on the line . Now use two point form . |
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| 3. |
The equation of the line passing through point `(-3,-7)` and making an intercept of 10 units on X-axis can be `"_______"`.A. 4x + 3y = -9B. 8x - 3y = 80C. 7x - 13y - 70 = 0D. 7x + 3y - 70 = 0 |
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Answer» Correct Answer - C If x-intercept is 10 , then the line passing through (10 , 0) . Now , use the two point form . |
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| 4. |
The equation of the line whose x-intercept is 5 , and which is parallel to the line joining the points (3,2) and `(-4, -1)` is `"_______"`.A. `4x + 7y - 20 = 0 `B. 3x - 7y + 3 = 0C. 3x + 2y + 15 = 0D. 3x - 7y - 15 = 0 |
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Answer» Correct Answer - D (i) Find m, then use slope-point form. (ii) Find the equation of the line passing through the given points. (iii) Any line parallel to `ax + by + c_(1)= 0` is `ax + by + c_(2) = 0` (iv) The required line `ax + by + c_(2) = 0` passes through `(5,0)`. |
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| 5. |
The line joining the points (2m + 2 , 2m) and (2m + 1, 3) passes through (m +1 , 1) , if the values of m are `"_______"`.A. `5 , -(1)/(5)`B. `1 , -1`C. `2 , - (1)/(2)`D. `3 , -(1)/(3)` |
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Answer» Correct Answer - C (i) The three points are collinear. (ii) Given points `A, B and C are collinear. (iii) Use, slope of Ab = slope of Ac and find m. |
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| 6. |
Find the area of the triangle formed by the line 3x - 4y + 12 = 0 with the coordinate axes .A. 6 `"units"^(2)`B. `12 "units"^(2)`C. `1 "units"^(2)`D. `36 "units"^(2)` |
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Answer» Correct Answer - A (i) Area `= 1/2|(c^(2))/(ab)|`, when the equation of the line is `ax+by + c = 0`. (ii) if a and b are x- and y-intercepts, then the area of the triangle formed by the line with coordination axes is `|(ab)/(2)|` , |
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| 7. |
Write the coodinates of midpoint of the segment joining (4,5) and (12,15). |
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Answer» The coordinates of midpoint of the segment joining (4,5) and (12,15) is(8,10) |
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| 8. |
If (2, p) is the midpoint of the line segment joining the points A(6, -5) and B(-2, 11), find the value of p. |
| Answer» Correct Answer - p = 3 | |
| 9. |
Find the lengths of the medians AD and BE of `Delta ABC` whose vertices are A(7, -3), B(5, 3) and C(3, -1). |
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Answer» Correct Answer - AD = 5 units, BE = 5 units Midpoint of BC is `D((5+3)/(2), (3-1)/(2)), i.e., D(4, 1)` Midpoint of AC is `E((7+3)/(2), (-3-1)/(2)), i.e., E (5, -2)` `AD = sqrt((7-4)^(2) + (-3-1)^(2)) = sqrt(3^(2) + (-4)^(2)) = sqrt(25) = 5` units. `BE = sqrt((5-5)^(2) + (3+2)^(2)) = sqrt(0+25) = sqrt(25) = 5` units. |
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| 10. |
Find the point on x-axis which is equidistant from points A(-1, 0) and B(5, 0). |
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Answer» Correct Answer - P(2, 0) Let the required point be P(x, 0). Then, `PA^(2) = PB^(2) rArr (x+1)^(2) + (0-0)^(2) = (x-5)^(2) + (0-0)^(2)` `rArr x^(2) +1 +2x = x^(2) + 25 -10x` `rArr 12x = 24 rArr x = 2` `therefore` the required point is P(2, 0). |
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| 11. |
Find the fourth vertex of the parallelogram whose three consecutive vertices are (8,8) , (6,1) and (-1,1) . |
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Answer» Let the three vertices of the parallelogram be A(8,8) , B(6,1) and C(-1,1) , then fourth vertex D( x, y) is given by `D(x , y) = (x_(1) - x_(2) + x_(3) , y_(1) - y_(2) + y_(3))` `= ( 8 - 6 - 1 , 8 - 1 + 1)` = `(1,8)` . Hence , the fourth vertex is D(1,8) . |
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| 12. |
Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, -3). |
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Answer» Correct Answer - `3:5` Let the required ratio be k : 1. Then, `(-3k + 5)/(k+1) = 2 rArr -3k +5 = 2k + 2 rArr 5k = 3 rArr k = (3)/(5)` Required ratio ` = ((3)/(5) : 1) = 3 : 5` |
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| 13. |
If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2:3 then find the value of k. |
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Answer» Correct Answer - `k = (16)/(5)` `k = ((2 xx 5 + 3 xx 2))/((2+3)) = (16)/(5)` |
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| 14. |
Find the coordinates of the midpoint of the line segment joining the points A(-5, 4) and B(7, -8). |
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Answer» Let M(x, y) be the midpoint of AB. Then, `x = ([(-5)+7])/(2) = 1 "and" y = ([4+(-8)])/(2) = 2.` Hence, the required point is M(1, -2). |
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| 15. |
If threeconsecutive vertices of a parallelogram are `(1, -2), (3, 6)`and `(5, 10)`, find itsfourth vertex. |
| Answer» Correct Answer - D(3, 2) | |
| 16. |
The midpoint of segment AB is P(0, 4). If the coordinates of B are (-2, 3), then the coordinates of A areA. (2, 5)B. (-2, -5)C. (2, 9)D. (-2, 11) |
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Answer» Correct Answer - A Let the point A be (a, b). Then, `(a +(-2))/(2) = 0 "and" (b+3)/(2) = 4` `rArr a -2 = 0 "and" b = 8-3 rArr a = 2, b = 5.` `therefore ` the point A is (2, 5). |
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| 17. |
(i) Find the coordinates of the point which divides the join of A(-1, 7) and B(4, -3) in the ratio 2:3. (ii) Find the coordinates of the point which divides the join of A(-5, 11) and B(4, -7) in the ration 7:2. |
| Answer» Correct Answer - (i) (1, 3) (ii) (2, -3) | |
| 18. |
Find the ratio in which the point (-3, k) divides the join of A(-5, -4) and B(-2, 3). Also, find the value of k. |
| Answer» Correct Answer - `2:1, k = (2)/(3)` | |
| 19. |
Find the area of `Delta ABC` whose vertices are : (i) A(1, 2), B (-2, 3) and C(-3, -4) (ii) A(-5, 7), B(-4, -5) and C(4, 5) (iii) A(3, 8), B(-4, 2) and C(5, -1) (iv) A(10, -6), B(2, 5) and C(-1, 3) |
| Answer» Correct Answer - (i) 11 sq units (ii) 53sq units (iii) 37.5 sq units (iv) 24.5 sq units | |
| 20. |
A line intersects the Y- axis at the points P and Q,, respectively. If (2,-5) is the mid- point of PQ , then the coordinates of P and Q are , respectively |
| Answer» Correct Answer - P(0, -10), Q(4, 0) | |
| 21. |
The area of `Delta ABC` with vertices A(a, 0), O(0, 0) and B(0, b) in square units isA. abB. `(1)/(2)ab`C. `(1)/(2)a^(2)b^(2)`D. `(1)/(2)b^(2)` |
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Answer» Correct Answer - B Here, base = a units and height = b units. `therefore "area" = ((1)/(2) xx "base" xx "height")` `=(1)/(2)` ab sq units. |
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| 22. |
Find the value of k so that the area of the triangle with vertices A(k+1, 1), B(4, -3) and C(7, -k) is 6 square units. |
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Answer» Correct Answer - k = 3 Let `A(x_(1) = k + 1, y_(1) =1), B(x_(2) = 4, y_(2) = -3) " and "C(x_(3) = 7, y_(3) = -k)` be the vertices of `Delta ABC`. Then, `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)|(k+1)(-3+k) + 4(-k-1) + 7(1+3)|` ` = (1)/(2)|(k^(2) -6k + 21)| = (1)/(2)|(k^(2) -6k + 9) +12|` ` = (1)/(2) |(k-3)^(2) + 12| = (1)/(2)[(k-3)^(2) + 12].` `therefore (1)/(2)[(k-3)^(2) +12] = 6 rArr (k-3)^(2) + 12 = 12 rArr (k-3)^(2) = 0` Hence, k = 3. |
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| 23. |
In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(-6, 9)? |
| Answer» Correct Answer - `3:4` | |
| 24. |
In the given figure P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then, y equals A. 2B. 4C. `-4`D. `(-5)/(2)` |
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Answer» Correct Answer - C Q(3, y) divides AB in the ratio 2:1. So, Q is `((2 xx 1 + 1 xx 7)/(2+1), (2 xx (-5) + 1 xx (-2))/(2+1)), i.e., (3, -4).` Hence, y = -4. |
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| 25. |
The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2:1 isA. (2, 4)B. (3, 5)C. (4, 2)D. (5, 3) |
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Answer» Correct Answer - B Coordinates of P are `(((2 xx 4 + 1 xx 1))/(2+1), ((2 xx 6 +1 xx 3))/(2+1)) = (3, 5)` |
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| 26. |
Two vertices of `Delta ABC` are A(-1, 4) and B(5, 2) and its centroid is G(0, -3). Then, the coordinates of C areA. (4, 3)B. (4, 15)C. (-4, -15)D. (-15, -4) |
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Answer» Correct Answer - C Let the vertex C be C(x, y). Then, `(-1+5+x)/(3) = 0 "and" (4+2+y)/(3) = -3 rArr x +4 =0 "and" 6+y =-9` `therefore x = -4 "and" y = -15.` So, the coordinates of C are (-4, -15). |
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| 27. |
(i) If the vertices of `Delta ABC` be A(1, -3), B(4, p) and C(-9, 7) and its area is 15 square units, find the value of p. (ii) The area of a triangle is 5sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is `((7)/(2), y)`, find the value of y. 1 |
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Answer» Correct Answer - (i) p = -3 or p = -9 (ii) y = `(13)/(2)` (i) Taking `A(x_(1) = 1, y_(1) = -3), B(x_(2) = 4, y_(2) = p) "and" C(x_(3) = -9, y_(3) = 7)`, we get ar`(Delta ABC)= 15` sq units. `rArr (1)/(2)|x_(1)(y_(2) - y_(3)) + x_(2) (y_(3) - y_(1)) + x_(3) (y_(1) -y_(2))|= 15` `rArr |1 * (p-7) + 4 * (7+3) -9 * (-3-p)|= 30` `rArr |(p-7) + 40 + 27 + 9p| = 30 rArr |10p + 60| = 30` `rArr 10p + 60 = 30 "or" 10p + 60 = -30 rArr p = -3 "or" p = -9.` |
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| 28. |
Find the coordinates of a point A, where AB is a diameter of a circle with centre C(2, -3) and the other end of the diameter is B(1, 4). |
| Answer» Correct Answer - A(3, -10) | |
| 29. |
The line segment joining A(-2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C. |
| Answer» Correct Answer - C(2, 6) | |
| 30. |
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is `(2, 3)` and `B` is `(1, 4)`. |
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Answer» ((x+1)/2,(y+4)/2)=(2,-3) (x+1)/2=2,(y+4)/2=-3 x=3,y=-10 |
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| 31. |
Find the centroid of `Delta ABC` whose vertices are A(-3, 0), B(5, -2), and C(-8, 5). |
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Answer» Here, `(x_(1) = -3, y_(1) = 0), (x_(2) = 5, y_(2) = -2) "and" (x_(3) = -8, y_(3) = 5).` Let G(x, y) be the centroid of `Delta ABC`. Then, `x = (1)/(3) (x_(1) +x_(2) + x_(3)) = (1)/(3) (-3+5-8) = -2,` `y = (1)/(3) (y_(1) + y_(2) + y_(3)) = (1)/(3)(0-2+5) = 1.` Hence, the centroid of `Delta ABC` is G(-2, 1). |
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| 32. |
ABCD is a rectangle formed by the points `A(1, 1),``B(1, 4),`` C(5, 4)` and `D(5, 1)`. P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? A rectangle? or a rhombus? Justify your answer. |
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Answer» Here,Coordinates of `P = ((-1-1)/2,(-1+4)/2) = (-1,3/2)` Coordinates of `Q = ((-1+5)/2,(4+4)/2) = (2,4)` Coordinates of `R = ((5+5)/2,(4-1)/2) = (5,3/2)` Coordinates of `S = ((5-1)/2,(-1-1)/2) = (2,-1)` So, `PQ = sqrt(3^2+(5/2)^2)=sqrt61/2` `QR = sqrt(3^2+(-5/2)^2) = sqrt61/2` `RS = sqrt((-3)^2+(-5/2)^2) = sqrt61/2` `SP = sqrt((3)^2+(-5/2)^2) = sqrt61/2` It means, all sides are `equal`. Now, we check diagonals. `PR = sqrt(6^2+0) = 6` `QS = sqrt(0+(-5)^2) = 5` It means, diagonals are not equal. As all sides are equal and diagonals are not equal, `PQRS` is a rhombus. |
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| 33. |
A (-2,-1) , B (1,0), C (4,3) and D (1,2) are the vertices of ` square ` ABCD then Using midpoint formula , find the coordinates of midpoints of join of B and D . |
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Answer» `square ABCD ` is parallelogram, since diagonals of quadriateral bisect each other |
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| 34. |
A(4,2), B(6,5) and C(1,4) are the vertices of `triangle ABC ` Find coordinates of points P on AD such that ` AP : PD= 2: 1 ` . |
| Answer» `P((11)/(3),(11)/(3))` | |
| 35. |
`A(4, 2), B(6,5)`and `C(1, 4)`are thevertices of ` A B C`. Themedian from `A`meets `B C`in `D`. Find thecoordinates of the point `D`. |
| Answer» `D((7)/(2),(9)/(2))` | |
| 36. |
The line ax + by + c = 0 meets Y-axis at `"_______"` point . |
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Answer» Correct Answer - `(0 , (-a)/(b))` |
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| 37. |
Show that points A(-1, 0) , B(-2,1) , C(1,3) and D(2,2) form a parallelogram . |
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Answer» Given A(-1,0) , B(-2,1) , C(1,3) and D(2,2) . `AB = sqrt((-2+1)^(2) + (1-0)^(2)) = sqrt2` units BC = `sqrt((1-(-2))^(2) + (3-1)^(2)) = sqrt(13)` units `CD = sqrt((2-1)^(2) + (2-3)^(2)) = sqrt2` units `DA = sqrt((2-(-1))^(2) + (2-0)^(2)) = sqrt(13)` units `AC = sqrt((1-(-1))^(2) + (3-0)^(2)) = sqrt(13)` units BD = `sqrt((2-(-2))^(2) + (2-1)^(2)) = sqrt(17)` units Clearly , AB = CD , BC = DA and AC `ne` BD . That is the opposite sides of the quadrilateral are equal and diagonals are equal . Hence , the given points form a parallelogram . |
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| 38. |
Find the area of the circle whose centre is (-1, -2) and `(3,4)` is a point on the circle . |
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Answer» Let the centre of the circle of the circle be A(-1,-2) , and the point on the circumference be B(3,4) . Radius of circle = AB `= sqrt((3-(-1))^(2) + (4-(-2))^(2)) = sqrt(52)` units . `therefore` The area of the circle = `pir^(2)` = `pi(sqrt(52))^(2) = 52 pi ` sq. units. |
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| 39. |
The lines `a_(1)x + b_(1)y + c_(1) = 0` and `a_(2)x + b_(2)y + c_(2) = 0` are perpendicular to each other , then `"_______"`. |
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Answer» Correct Answer - `a_(1) a_(2) + b_(1) b_(2) = 0` |
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| 40. |
The joint of intersection of X-axis and Y-axis is `"_______"` |
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Answer» Correct Answer - origin (or) (0,0) |
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| 41. |
The points (0,0), (0,4) and (4,0) form a/an `"_______"` triangle . |
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Answer» Correct Answer - right -angled isoscles triangle |
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| 42. |
Find the circum-centre and the circum-radius of a triangle ABC formed by the vertices A(2,-2) , B (-1,1) and C(3,1). |
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Answer» Let S(x,y) be the circum-centre of `Delta`ABC . `therefore SA^(2) = SB^(2) = SC^(2)` Consider `SA^(2) = SB^(2)` `implies (x-2)^(2) + (y+2)^(2) = (x +1)^(2) + (y-1)^(2)` `x^(2) - 4x + 4 + y^(2) + 4y + 4 = x^(2) + 2x + 1 + y -2y + 1` `-4x + 4y + 8 = 2x -2y + 2` `6x - 6y - 6 = 0` `x - y - 1= 0" " (1)` `SB^(2) = SC^(2) ` `implies (x+1)^(2) + (y-1)^(2) = (x-3)^(2) + (y-1)^(2)` `x^(2) + 2x + 1 + y^(2) - 2y + 1 = x^(2) - 6x + 9 + y - 2y +1` `2x - 2y + 2 = -6x - 2y + 10` `8x - 8 = 0` `implies x -1`. Substituting x = 1 in Eq. (1) , we get y = 0 . `therefore` The required circum-centre of `DeltaABC` is (1,0). Circum-radius , SA = `sqrt((2-1)^(2) + (-2-0)^(2)) = sqrt5` units. |
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| 43. |
If (2x + 3y + 1) + `lambda` ( x - 2y - 3) = 0 represents the equation of a horizontal line , then find the value of `lambda` . |
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Answer» Correct Answer - `lambda = -2` |
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| 44. |
The line 2y + 3 = 0 and x = 3 intersect at `"______"`. |
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Answer» Correct Answer - `[3 - (3)/(2)`] |
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| 45. |
Find the equation of a line passing through point (-2,3) and perpendicular to 7x + 2y + 3 = 0. |
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Answer» Here, `(x_(1) , y_(1)) = (-2,3) , a = 7` and b = 2 . `therefore` Equation of the line perpendicular to 7x + 2y + 3 = 0 and passing through `(-2,3)` is `b(x - x_(1)) - a(y_ - y_(1)) = 0` . That is , 2(x+2) - 7(y-3) = 0 `implies 2x - 7y + 25 = 0` . Hence the required equation of the line is 2x - 7y + 25= 0 . |
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| 46. |
The circum-centre of the triangle formed by points O (0,0) , A(6,0) and B (0, 6) is `"_______"`.A. (3,3)B. (2,2)C. (1,1)D. (0,0) |
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Answer» Correct Answer - A The circum-centre of right-angled triangle is the mid-point of its hypotenuse. |
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| 47. |
The line x - 2y + 3 = 0 , 3x - y = 1 and kx - y + 1 = 0 are concurrent . Find k .A. 1B. `(1)/(2)`C. `(3)/(2)`D. `(5)/(2)` |
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Answer» Correct Answer - A Solve the first two equations and substitute `(x,y)` in the third equation and evaluate k. |
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| 48. |
The lines `3x - y + 2= 0 ` and x + 3y + 4 = 0 intersect each other in the line `"_______"`.A. 1st quadrantB. 4th quadrantC. 3rd quadrantD. 2nd quadrant |
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Answer» Correct Answer - C Find the corrdinates of the point of intersection. |
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| 49. |
See Fig. 3.11 and complete the following statements:(i) The abscissa and the ordinate of the point B are _and _ Hence, the coordinates of B are (__,__).(ii) The x–coordinate and the y–coordinate of the point M are _ and _ respectively. Hence, the coordinates of M are (__,__).(iii) The x–coordinate and the y–coordinate of the point L are _ and _ respectively. Hence, the coordinates of L are (__,__).(iv) The .r–coordinate and the y–coordinate of the point S are _ and _ respectively. Hence, the coordinates of S are (__,__). |
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Answer» (i) `4,3,(4,3)` (ii) `-3,4,(-3,4)` (iii)` -5,-4, (-5,-4)` (iv)` 3,-4 , (3,-4)` answers |
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| 50. |
Find the quadrant in which the line 2x + 3y - 1 = 0 and 3x + y - 5 = 0 intersect each other .A. 1st quadrantB. 4th quadrantC. 3rd quadrantD. 2nd quadrant |
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Answer» Correct Answer - D Find the point o intersection of the given lines, then decide. |
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