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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.
| 151. |
The line segment joining the points (-3,-4) and (1, -2) is divided by Y-axis in the ratio.A. `2:3`B. `3:2`C. `3:1`D. 1:3` |
| Answer» Correct Answer - C | |
| 152. |
Find the values of k, if the points A (k+1,2k) ,B (3k,2k+3) and C (5k-1,5k) are collinear. |
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Answer» We know that , if three points are collinear , then the area of triangle formed by these points is zero. Since, the points A (k+1,2k) , B (3k,2k+3) and C (5k-1,5k) are collinear. Then , area of `triangle ABC =0` `rArr (1)/(2)[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2)]=0` Here , `x_(1)=k+1,x_(2)=3k,x_(3)=5k-1and y_(1)=2k+3,y_(3)=5k` `rArr(1)/(2)[(k+1)(-3k+3)+3k(3k)+(5k-1)(2k-3)]=0` `(1)/(2)[(k+1)(-3k+3)+3k(3k)+(5k-1)(2k-2k-3)]=0` `(1)/(2)[-3k^(2)+3k-3k+3+9k^(2)=15k+3]=0` `rArr(1)/(2)(6k^(2)-15k+6)=0` [multiply by 2] `rArr6k^(2)-15k+6=0` [by factorisation method] `2k^(2)-5k+2=0` [divide by 3] `rArr2k^(2)-4k-k+2=0` `2k(k-2)-(k-2)=0` `rArr(k-2)(2k-1)=0` If `k-2=0` , then `k=2` If `2k-1=0, then k=(1)/(2)` `:. k=2,(1)/(2)` Hence , the required values of k are and `(1)/(2)`. |
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| 153. |
If thepoints `A(-1, -4), B(b , c)`and `C(5, -1)`arecollinear and `2b+c=4`, find thevalues of `b`and `c`. |
| Answer» Answer: ` b= 3, c = -2` | |
| 154. |
For what value of x are the points A(-3, 12), B(7, 6) and C(x, 9) collinear? |
| Answer» Correct Answer - x =2 | |
| 155. |
Find the product of intercepts of the line 3x + 8y - 24 = 0.A. 8B. 24C. 3D. 12 |
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Answer» Correct Answer - B `3x+8y = 24` `rArr x/8 + y/3 = 1` `rArr` intercepts are 8 and 3 and their product is `8 xx 3 = 24`. |
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| 156. |
Find the equation of a line passing through the point (-3,2) and parallel to line 4x - 3y - 7 = 0 . |
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Answer» Here `(x_(1) , y_(1)) = (-3 , 2) , a = 4` and b = -3 `therefore` Equation of the line passing through P(-3,2) and parallel to 4x - 3y - 7 = 0. `a(x-x_(1)) + b(y-y_(1)) = 0` `implies 4(x + 3) - 3(y-2) = 0` `implies 4x - 3y + 18 = 0` . Hence . the equation of the required line is 4x - 3y + 18 = 0. |
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| 157. |
If the roots of the quadratic equation `x^(2) - 7 x+ 12 = 0` are intercepts of a line , then the equation of the line can be `"_______"`.A. 2x + 3y = 6B. 4x + 3y = 12C. 4x + 3y = 6D. 3x +4y = 6 |
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Answer» Correct Answer - B Find the roots of the given equation , then use intercepts form of line . |
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| 158. |
Find the value of k , if points (-2,5) , (-5 ,-10) and (k , -13) are collinear .A. `(5)/(28)`B. `(-28)/(5)`C. 28D. 5 |
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Answer» Correct Answer - B `(-2,-5) (-5,-10)` and `(k,-13)` arec collinear `rArr (-10-5)/(-5+2) = (-13+10)/(k+5)` `rArr (-15)/(-3) = (-3)/(k+5)` `rArr 5(k+5) = -3` `rArr k +5 = (-3)/(5)` `rArr k = (-3)/(5) - 5 = (-28)/(5)` `:. K = (-28)/(5)`. |
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| 159. |
Find the area of the triangle formed by joining the mid–points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. |
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Answer» `D=((2/2),(1-1)/2)=(1,0)` `E=((2/2),(1+3)/2)=(1,2)` `F=(0,(3-1)/2)=(0,1)` Area of `/_DEF=1/2(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))` =`1/2(1(2-1)+1(1-0)+0)` =1 Area of `/_ABC=1/2(0+2(3+1)+0)` =4 ratio=`(area/_DEF)/(area/_ABC)=1/4`. |
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| 160. |
Find the intercepts made by the line 3x - 2y - 6 = 0 on the coordinate axes . |
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Answer» Correct Answer - x - intercept (a) = 2 y-intercept (b) = `-3` |
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| 161. |
The equation of the line making intercepts of equal magnitude and opposite signs , and passing through the point (-3,-5) is `"_______"`.A. `x- y = 2`B. `2x + y = -4`C. 3x + 3y = 6D. `x - y = -10` |
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Answer» Correct Answer - A Use `(x)/(a) + (y)/(b) = 1 ` then a = `-b`. |
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| 162. |
Find the product of intercepts made by the line 7x - 2y - 14 = 0 with coordinate axes .A. `-7`B. 2C. 14D. `-14` |
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Answer» Correct Answer - D `7x - 2y - 14 = 0 rArr 7x - 2y = 14` `rArr x/2 - y/7 = 1 rArr x/2 + (y)/((-7)) = 1`. `:.` intercepts are 2 and -7 and their product `= 2xx (-7) = - 14` |
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| 163. |
The equation of the line making equal intercepts and passing through the point (-1,4) is `"_______"`A. x - y = 3B. x + y + 3 = 0C. x + y = 3D. x - y + 3 = 0 |
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Answer» Correct Answer - C Equation of the line making equal intercept is of the form , x + y = a . |
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| 164. |
The sum of the reciprocals of the intercepts of a line is `(1)/(2)` , then the line passes through the point is `"________"`. (a) (1, 1) `" " ` (b) (2,1) `" " (c) ((1)/(4) , (1)/(4))` (d) (2,2) |
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Answer» (i) Use `(x)/(a) + (y)/(b) = 1`. (ii) Solve , `(1)/(a) + (1)/(b) = (1)/(2)` and get the relation between a and b . (iii) Use the formula `(x)/(a) + (y)/(b) = 1`. |
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| 165. |
Find the mid-point of the line segment joining the points (2,-6) and (6,-4) . |
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Answer» Let A(2,-6) and B(6,-4) be the given points and M be the mid-point of AB . Then , `M = ((x_(1) + x_(2))/(2) , (y_(1) + y_(2))/(2))` `= ((2+6)/(2), (-6 +(-4))/(2)) = (4 , -5)`. Hence , the mid point of AB is `(4,-5)`. |
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| 166. |
Find the equation of the line passing through `(-5,11)` and making equal intercepts , but opposite in magnitude on the coordinate axes. |
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Answer» Correct Answer - x - y + 16 = 0 |
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| 167. |
The area of triangle formed by the line y = mx + c with the coordinate axes is `"_______"`. |
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Answer» Correct Answer - `(1)/(2) |(c^(2))/(m)|`sq . Units |
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| 168. |
The equation of a line passing through P (3,4) such that P bisects the part of it intercepted between the coordinate axes is `"_______"`.A. 3x + 4y = 25B. 4x + 3y = 24C. x - y = -1D. x + y = 7 |
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Answer» Correct Answer - B (i) P is the mid-point of the line joining the inter-cepts. Find the intercepts using the mid-point formula. (ii) Let the line cut coodinate axes at `A(a,0)` and `B(0,b)`. (iii) Using the above data find a and b, then the equation of line, i.e, `x/a + y/b = 1`. |
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| 169. |
Find the ratio in which the point P (11,y) divides the line segment joining the points `A(15,5) and B(9,20).` Also find the value of y. |
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Answer» Let the required ratio be k: 1. Then, by the section formula, the coordinates of P are `P((9k +15)/(k +1), (20k+5)/(k+1)).` But, this point is given as P(11, y). `therefore (9k +15)/(k+1) = 11 rArr 9k +15 = 11k + 11 rArr 2k = 4 rArr k = 2.` So, the required ratio is 2:1. Putting k = 2 in P, we get `y = (20 xx 2 +5 )/((2+1)) = (45)/(3) = 45.` Hence, y = 15. |
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| 170. |
If G(-2, 1) is the centroid of a `DeltaABC` and two of its vertices are A(1, -6) and B(-5, 2), find the third vertex of the triangle. |
| Answer» Correct Answer - C(-2, 7) | |
| 171. |
If (-4, 0) and (4,0) are two vertices of an equilateral triangle, find the coordinates of its third vertex. |
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Answer» Let `A(x,y)` is the coordinate of the third point. We are given `B(-4,0)` and `C(4,0)`. Here, `BC = sqrt(4-(-4)^2+(0-0)^2) = 8` `AB = sqrt((x-(-4))^2+(y-0)^2) = sqrt((x+4)^2+y^2)` `AB = sqrt((x-4))^2+(y-0)^2) = sqrt((x-4)^2+y^2)` As given triangle is an equilateral triangle. `:. AB = AC = BC` ` sqrt((x+4)^2+y^2) = sqrt((x-4)^2+y^2) ` `=>(x+4)^2+y^2 = (x-4)^2+y^2 ` `=>(x+4)^2 = (x-4)^2` `=>x^2+16+8x = x^2+16-8x` `=>16x = 0` `=>x = 0` Now, `AB = BC` `:. sqrt((x+4)^2+y^2) = 8` `=> (x+4)^2+y^2 = 64` `=>(0+4)^2+y^2 = 64` `=>y^2 = 64-16` `=> y = sqrt48 = 4sqrt3` So, coordinates of third vertex will be `A(0,4sqrt3)`. |
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| 172. |
Three vertices of a rectangle ABCD are A(3, 1), B(-3, 1) and `C(-3, 3)`. Plot these points on a graph paper and find the coordinates of the fourth vertex D. Also, find the area of rectangle ABCD. Hint Coordinates of D are (3, 3) |
| Answer» Correct Answer - D(3, 3), q sq units | |
| 173. |
23. Write the coordinates of the vertices of a rectangle whose length and breadth are one 5 and 3 units respectively, one vertex at origin, the longer side on the x-axis and of the vertices in the third quadrant. |
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Answer» Given , length of a rectangle=5 units and breadth of a rectangle=3 units One vextex is at origin i.e., (0,0) and one of the other vertices vertices lies in III quadrant. So, the length of the rectangle is 5 units in the negative direction of X-axis and then vertex is A(-5,0) . Also, the breadth of the rectangle is 3 units in the negative direction of Y-axis and then vertex is C(0,-3). The fourth vextex B is (-5,-3). |
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| 174. |
P(-5,-3), Q(-4,-6), R(2,-3) and S (1,2) are the vertices of a quadrilateral PQRS, find its area |
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Answer» Area of `/_PQR` `=1/2(x1(y2-y3)+x2(y3+y1)+x3(y1-y2)` `=1/2(-5(-6+3)+(-4)(-3+3)+2((-3)+6)` `=1/2(+15+0+6)` `=21/2unit^2` Area of `/_PQR` `=1/2(-5(-3-2)+(2)(2+3)+1((-3)+3)` `=1/2(25+10+0)` `=35/2 unit^2` Area of quad.=Area of `/_PQR`+Area of `/_PRS` `=21/2+35/2` `=56/2` `=28Unit^2` |
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| 175. |
lf G be the centroid of a triangle ABC and P be any other point in the plane prove that `PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3GP^2` |
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Answer» `G ` is intersection of median `D` is mid[point of DC `G(0,0)` Coordinates of D=`((c + e)/2, (d+f)/2)` centroid of triangle G=`((a+c+e)/3 , (b+d+f)/3)` `(a+c+e)/3 = 0& (b+d+f)/3 = 0` `e= -(a+c) & f= - (b+d)` `PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^3 + 3GP^2` `[(x-a)^2 + (y-b)^2] + [(x-c)^2+ (y-d)^2] + [(x+a+c)^2 + (y+b+d)^2]` `= [(0-a)^2 + (0-b)^2] + [(0-c)^2 + (0-d)^2] + [(a+c)^2 + (b+d)^2] + 3[(x-0)^2 + (y-0)^2]` from LHS `x^2 + a^2 - 2ax + y^2 + b^2 - 2by + x^2 + c^2 - 2cx + y^2 + d^2 = 2dy + x^2 + a^2 + c^2 ` `= 3(x^2 + y^2) + 2(a^2 + b^2 +c^2 + d^2+ c^2 + d^2 + ca+bd)` from RHS `a^2 + b^2 + c^2 + d^2 + a^2 + c^2 + 2ac + b^2 + d^2 + 2bd + 3(x^2 + y^2)` `= 3[x^2 + y^2] + 2(a^2 + b^2 + c^2 + d^2 + ac+bd)` LHS=RHS Hence proved |
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| 176. |
The points P(0, 6), Q(-5, 3) and R(3, 1) are the vertices of a triangle which isA. equilateralB. isoscelesC. scaleneD. right angled |
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Answer» Correct Answer - D `PQ^(2) = (-5-0)^(2) + (3-6)^(2) = (-5)^(2) + (-3)^(2) = 25 +9 = 34` `QR^(2) = (3+5)^(2) + (1-3)^(2) = 8^(2) + (-2)^(2) = 64+ 4 = 68` `PR^(2) = (3-0)^(2) + (1-6)^(2) = 3^(2) + (-5)^(2) = (9+25) = 34` `therefore PQ^(2) + PR^(2) = QR^(2)` Hence, `Delta PQR` is right-angled. |
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| 177. |
Find the value of k if the points (2,3) , (5,k) and (6,7) are collinear .A. k = 4B. k = 6C. `k = (-3)/(2)`D. `k = (11)/(4)` |
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Answer» Correct Answer - B Here, `(x_(1) = 2, y_(1) = 3), (x_(2) = 5, y_(2) = k) "and" (x_(3) = 6, y_(3) = 7)` Since the given points are collinear, we must have : `x_(1) (y_(2) - y_(3)) +x_(2)(y_(3)-y_(1)) + x_(3)(y_(1) -y_(2)) = 0` `rArr 2(k-7) + 5(7-3) +6(3-k) = 0 rArr 2k -14 + 20 +18 - 6k = 0` `rArr 4k = 24 rArr k = 6.` |
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| 178. |
If the points A(1,2), B(0,0) and C (a,b) are collinear, thenA. a= bB. a = 2bC. 2a = bD. a+b =0 |
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Answer» Correct Answer - C Here, `(x_(1) = 1, y_(1) = 2), (x_(2) = 0, y_(2) = 0)"and" (x_(3) = a, y_(3) = b)`. Since the given points are collinear, we have `x_(1) (y_(2) - y_(3)) +x_(2)(y_(3) -y_(1)) + x_(3)(y_(1) -y_(2)) = 0` `rArr 1 * (0-b) +0* (b-2) +a * (2-0) = 0 rArr -b+0+2a =0 rArr 2a = b`. |
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| 179. |
The area of a triangle with vertices A(3,0),B(7,0) and C(8,4) isA. 14sq unitsB. 28 sq unitsC. 8 sq unitsD. 6 sq units |
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Answer» Correct Answer - C Here, `(x_(1) = 3, y_(1) = 0), (x_(2) = 7, y_(2) = 0) "and" (x_(3) = 8, y_(3) = 4)` `therefore "area of "Delta ABC = (1)/(2) {x_(1) (y_(2)-y_(3)) +x_(2)(y_(3)-y_(1)) +x_(3)(y_(1) -y_(2))}` ` = (1)/(2) {3(0-4) + 7(4-0) + 8(0-0)} = (1)/(2){-12 +28 +0}` ` = ((1)/(2) xx 16) = 8` sq units. |
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| 180. |
If thedistance between the points `(4, p)`and `(1, 0)`is 5, then `p=` ?A. p = 4 onlyB. p = -4 onlyC. `p = +-4`D. p = 0 |
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Answer» Correct Answer - C `(4-1)^(2) + (p-0)^(2) = 5^(2) rArr p^(2) = (25-9) = 16 rArr p = +-4.` |
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