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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. |
Find the angle between the lines whose joint equation is `2x^2-3xy+y^2=0`A. `tan^(-1)(sqrt(3))`B. `cot^(-1)(sqrt(3))`C. `cot^(-1)(3)`D. `cos^(-1)(3)` |
| Answer» Correct Answer - C | |
| 152. |
Joint equation of two lines through (2,-1) parallel to two lines `2x^(2)-3xy-9y^(2)=0` isA. `2x^(2)-3xy+9y^(2)-5x-24y-7=0`B. `2x^(2)-3xy-9y^(2)-5x-24y-7=0`C. `2x^(2)+3xy-9y^(2)-5x-24y-7=0`D. `2x^(2)+3xy-9y^(2)-5x-24y-7=0` |
| Answer» Correct Answer - C | |
| 153. |
Joint equation of two lines both parallel to X-axis, and each at a distance of 2 units from it isA. `x^(2)-4=0`B. `y^(2)-4=0`C. `x^(2)-y^(2)=4`D. `y^(2)+4=0` |
| Answer» Correct Answer - B | |
| 154. |
Joint equation of co-ordinates axes, in a plane isA. `x^(2)-y^(2)=0`B. `x^(2)+y^(2)=1`C. `xy=0`D. `xy=x+y` |
| Answer» Correct Answer - C | |
| 155. |
Joint equation of lines bisecting angles between co-oridnates axes isA. `x^(2)+y^(2)=0`B. `x^(2)-y^(2)=0`C. `x^(2)-2y^(2)=0`D. `x^(2)+y^(2)=1` |
| Answer» Correct Answer - B | |
| 156. |
Joint equation of lines, trisecting angles is second and fourth quadrant isA. `sqrt(3)(x^(2)+y^(2))-4xy=0`B. `sqrt(3)(x^(2)-y^(2))-4xy=0`C. `sqrt(3)(x^(2)+y^(2))+4xy=0`D. `4(x^(2)+y^(2))+sqrt(3)xy=0` |
| Answer» Correct Answer - C | |
| 157. |
Joint equation of lines, trisecting angles in first and third quadant isA. `sqrt(3)(x^(2)-y^(2))-4x=0`B. `sqrt(3)(x^(2)-y^(2))+4xy=0`C. `sqrt(3)(x^(2)+y^(2))+4xy=0`D. `sqrt(3)(x^(2)+y^(2))-4xy=0` |
| Answer» Correct Answer - D | |
| 158. |
The equation `8x^2 +8xy +2y^2+26x +13y+15=0` represents a pair of straight lines. The distance between them isA. `(7)/(sqrt5)`B. `(7)/(2sqrt5)`C. `sqrt((7)/(5))`D. none of these |
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Answer» Correct Answer - B The distance between the parallel straight lines given by `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0` is given by `d=2sqrt((g^(2)-ac)/(a(a+b)))` Here, a = 8, b= 2 c = 15, g = 13. `therefore" Required distance = 2"sqrt((169-120)/(80))=2xx(7)/(4sqrt5)=(7)/(2sqrt5)` |
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| 159. |
The gradient of one of the lines given by `ax^(2)+2hxy+by^(2)=0` is twice that of the other, thenA. `h^(2)=ab`B. `h=a+b`C. `8h^(2)=9ab`D. `9h^(2)=8ab` |
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Answer» Correct Answer - C Let m and be the gradients of the lines given by `ax^(2)+2hxy+by^(2)=0`. Then, `m+2m=(-2h)/(b)and mxx2m=(a)/(b)` `rArr" "m=-(2h)/(3b)and m^(2)=(a)/(2b)` `rArr" "(4h^(2))/(9b^(2))=(a)/(2b)=rArr 8h^(2)=9ab` |
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| 160. |
If the equation `x^2+(lambda+mu)x y+lambdau y^2+x+muy=0`represents two parallel straight lines, then prove that `lambda=mudot`A. `lambda | mu=0`B. `lambda=4mu`C. `lambda-mu`D. none of these |
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Answer» Correct Answer - C The given equation will represent a pair of parallel lines, if `((lambda+mu)/(2))^(2)=lambdamu" "["Using ":h^(2)=ab]` `rArr" "(lambda-mu)^(2)rArr lambda=mu` |
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