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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. |
The lines represented by the equation `ax^(2)+2bxy+hy^(2)=0` are mutually perpendicular ifA. `a+b=0`B. `b+h=0`C. `h+a=0`D. `ah=-1` |
| Answer» Correct Answer - C | |
| 2. |
If acute angle between lines `x^(2)-2hxy+y^(2)=0` is `60^(@)` then `h=`A. `-2`B. `+-2`C. `2`D. `sqrt(3)` |
| Answer» Correct Answer - B | |
| 3. |
The angle between the lines `2x^2-7x y+3y^2=0`is`60^0`2. `45^0`3. `tan^(-1)(7//6)`4. `30^0`A. `60^(@)`B. `45^(@)`C. `tan^(-1)((7)/(6))`D. `30^(@)` |
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Answer» Correct Answer - B Here, `=2, b=3 and h=-(7)/(2)` Let `theta` be the angle between the lines represented by `2x^(2)-7xy+3y^(2)=0`. Then, `tan theta=(2sqrt((49)/(4)-6))/(5)=1rArrtheta=45^(@)` |
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| 4. |
Find the value of `a`for which the lines represented by `a x^2+5x y+2y^2=0`are mutually perpendicular. |
| Answer» The lines given by `ax^(2)+5xy+2y^(2)=0` are mutually perpendicular if `a+2=0,i.e.,a=-2`. | |
| 5. |
The two lines represented by `3a x^2+5x y+(a^2-2)y^2=0`are perpendicular to each other fortwo values of `a`(b) afor one value of `a`(d) for no values of `a`A. two values of aB. for one value of aC. for one value of aD. for no value of a |
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Answer» Correct Answer - A The pair of lines represented by the given equation will be perpendicular to each other, if `"Coefficient of "x^(2)+"Coefficient of "y^(2)=0` `rArr" "3a+a^(2)-2=0rArra=(-3pmsqrt(9+8))/(2)=(-3pmsqrt(17))/(2)` Thus, there are two values of a for which the lines represented by `2x^(2)-7xy+3y^(2)=0,` is |
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| 6. |
If acute angle between lines `ax^(2)+2hxy+by^(2)=0` is `(pi)/6`, then `a^(2)+14ab+b^(2)=`A. `4h^(2)`B. `8h^(2)`C. `12h^(2)`D. `16h^(2)` |
| Answer» Correct Answer - C | |
| 7. |
If acute angle between lines `ax^(2)+2hxy+by^(2)=0` is congruent to that between lines `2x^(2)-5xy+3y^(2)=0` and `k(h^(2)-ab)=(a+b)^(2)` then `k=`A. `-(10)^(2)`B. `(-10)^(2)`C. `-10`D. `10` |
| Answer» Correct Answer - B | |
| 8. |
If acute angle between lines `ax^(2)+2hxy+by^(2)=0` is `(pi)/4`, then `4h^(2)=`A. `a^(2)+4ab+b^(2)`B. `a^(2)+6ab+b^(2)`C. `(a+2b)(a+3b)`D. `(a-2b)(2a+b)` |
| Answer» Correct Answer - B | |
| 9. |
The lines `a^2x^2+bcy^2=a(b+c)xy` will be coincident , ifA. `a=b`B. `b=c`C. `c=a`D. `b^(2)=ac` |
| Answer» Correct Answer - B | |
| 10. |
Find the measure of the acute angle between the lines represented by `(a^(2) - 3b^(2)) x^(2) + 8abxy + (b^(2) - 3a^(2)) y^(2) = 0.`A. `(pi)/6`B. `(pi)/4`C. `(pi)/3`D. `(pi)/2` |
| Answer» Correct Answer - C | |
| 11. |
If the acute angle between the lines `ax^(2) + 2hxy + by^(2) = 0` is `60^(@)`, then show that `(a+3b)(3a+b) = 4h^(2)`.A. `h^(2)`B. `2h^(2)`C. `3h^(2)`D. `4h^(2)` |
| Answer» Correct Answer - D | |
| 12. |
Separate equations of lines jointly given by the equation `hxy+gx+(fh)/gh y+f=0` areA. `x=(-fh)/g,y=(-g)/h`B. `x=f/g,y=(-g)/h`C. `x=(-f)/h,h=(-g)/h`D. `fg=ch` |
| Answer» Correct Answer - A | |
| 13. |
If the equation `ax^(2)+hxy+by^(2)+4gx+6fy+4c=0` represents a pair of lines thenA. `4abc+4fgh=4.5af^(2)+4bg^(2)+h^(2)`B. `4abc+6fgh-9af^(2)+4bg^(2)+ch^(2)`C. `4abc+2fgh-9af^(2)+2bg^(2)+h^(2)`D. `4abc+12fgh-9af^(2)+4bg^(2)+2h^(2)` |
| Answer» Correct Answer - B | |
| 14. |
Does equation `x^2+2y^2-2sqrt(3)x-4y+5=0`satisfies the condition `a b c+2gh-af^2-bg^2-c h^2=0?`Does it represent a pair of straight lines? |
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Answer» Given equation is `x^(2)+2y^(2)-2sqrt(3)x-4y+5=0`. Here , `a=1,b=2,h=0,g=-sqrt(3),f=-2,c=5`. But , `abc+2gh-af^(2)-bg^(2)-ch^(2)=10+0-4-6=0`. The given equation can written as `(x-sqrt(3))^(2)+2(y-1)^(2)=0` Hence , it denotes only a point `P(sqrt(3),1)` but not a pair of straight lines . |
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| 15. |
Measure of angle bewtwen lines `(3+2sqrt(3))x^(2)-2xy-y^(2)=0` isA. `(pi)/2`B. `(pi)/3`C. `(pi)/4`D. `(pi)/6` |
| Answer» Correct Answer - C | |
| 16. |
The distance between the point of intersection of the two lines `2009x^(2)+2010xy+2011y^(2)=0` and the point (1,1) isA. 1B. `2`C. `sqrt(2)`D. `2+sqrt(3)` |
| Answer» Correct Answer - C | |
| 17. |
Which of the following pair of straight lines intersect at right angle?A. `2x^(2)=y(x+2y)`B. `(x+y)^(2)=x(y+3x)`C. `2y(x+y)=xy`D. `y= pm 2x` |
| Answer» Correct Answer - A | |
| 18. |
The equation `x^(2)+k_(1)y^(2)+k_(2)xy=0` represents a pair of perpendicular lines ifA. `k_(1)=-1`B. `k_(1)=2k_(2)`C. `2k_(1)=k_(2)`D. none of these |
| Answer» Correct Answer - A | |
| 19. |
If the equation `k^(2)x^(2)+10xy+3y^(2)-15x-21y+18=0` represents a pair of mutually perpendicular lines thenA. `k=5`B. `k=+-sqrt(2)`C. `k=3`D. `k` is not real |
| Answer» Correct Answer - D | |
| 20. |
The equations of a line which is parallel to the line common to the pair of lines given by `6x^(2)-xy-12y^(2)=0and15x^(2)+14xy-8y^(2)=0`and at a distance of 7 units from it iosA. `3x-4y=-35`B. `5x-2y=7`C. `3x+4y=35`D. `2x-3y=7` |
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Answer» Correct Answer - 3 We have `6x^(2)-xy-12y^(2)=0` or `(2x-3y)(3x+4y)=0` (1) and `15x^(2)+14xy-8y^(2)=0` or `(5x-2y)(3x+4y)=0` (2) Equation of the line common to (1) and(2) is `3x+4y=0` (3) Equation of any line parallel to (3) is `3x+4y=k` Since its distance from (3) is 7 ,we have `|(k)/(sqrt(3^(2)+4^(2)))|=7ork=+-35` |
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| 21. |
The equation of the lines parallel to the line common to the pair of lines given by `6x^(2)-xy-12y^(2)=0` and `15x^(2)+14xy-8y^(2)=0` and the sum of whose intercepts on the axes is 7, isA. `2x-3y=42`B. `3x+4y=12`C. `5x-2y=10`D. none of these |
| Answer» Correct Answer - B | |
| 22. |
If one of the lines `6x^(2)+exy+y^(2)=0` si `y+2x=0` then `c=`A. `-3`B. `-4`C. `-5`D. `5` |
| Answer» Correct Answer - D | |
| 23. |
If sum of slopes of lines `kx^(2)-10xy-9y^(2)=0` is live tme their product, then `k=`A. 2B. 1C. -2D. -1 |
| Answer» Correct Answer - A | |
| 24. |
If sum of slopes of lines `x^(2)+kxy-3y^(2)=0` is twice product of slopes then `k=`A. `-1`B. `-2`C. `1`D. `2` |
| Answer» Correct Answer - B | |
| 25. |
The equation of the lines which are parallel to the line common to the pair of the lines given by `6x^(2)-xy-12y^(2)=0` and `15x^(2)+14xy-8y^(2)=0` and at a distance of 7 units from it areA. `3x+4y=pm 35`B. `5x-2y= pm 7`C. `2x-3y= pm 7`D. none of these |
| Answer» Correct Answer - A | |
| 26. |
If `x^(2)-kxy+y^(2)+2y+2=0` denotes a pair of straight lines then k =A. 2B. `1//sqrt2`C. `2sqrt2`D. `sqrt2` |
| Answer» Correct Answer - D | |
| 27. |
If the lines represented by `x^(2)-2pxy-y^(2)=0` are rotated about the origin through an |
| Answer» Correct Answer - A | |
| 28. |
If the pair of lines `sqrt(3)x^2-4x y+sqrt(3)y^2=0`is rotated about the origin by `pi/6`in the anticlockwise sense, then find the equation of the pair in the newposition. |
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Answer» Correct Answer - `sqrt(3)x^(2)-xy=0` The given equation of pair of straight lines can be rewritten as `(sqrt(3)x-y)(x-sqrt(3y)=0` Their separate equation are `y=sqrt(3)xandy=(1)/(sqrt(3))x` or `y=tan60^(@)xandy=tan 30^(@)x` After rotation , the separate equations are `y=tan90^(@)xandy=tan 60^(@)x` or `x=0 and y=sqrt(3)x` Therefore , the combined equation of lines in the new position is `x(sqrt(3)x-y)=0orsqrt(3)x^(2)-xy=0` |
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| 29. |
If two of the lines given by the equation `ax^(3)+bx^(2)y+cxy^(2)+dy^(3)=0(a ne 0)` make complementary angles with x-axis in anticlockwise sense, thenA. `(a-c)=d(b-d)`B. `d(a-c)=a(d-b)`C. `a(a-c)=d(d-b)`D. none of these |
| Answer» Correct Answer - C | |
| 30. |
If the pair of straight lines `xy - x - y +1=0` & the line `ax+2y-3=0` are concurrent then `a =`A. -1B. 3C. 1D. 0 |
| Answer» Correct Answer - A | |
| 31. |
All chords of the curve `3x^2-y^2-2x+4y=0` which subtend a right angle at the origin, pass through the fixed point |
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Answer» The given curve is `3x^(2)-y^(2)-2x+4y=0` (1) Let `y=mx+c` be the chord of combined equations of lines joining the points of intersction of curve (1) and chord `y=mx+c`to the origin can be obtained by making the equation of curve homogeneous with the help of the equation of chord as `3x^(2)-y^(2)-2x((y-mx)/(c))+4((y-mx)/(c))=0` or `3cx^(2)-cy^(2)-2xy+2mx^(2)+4y^(2)-4mxy=0` or `(3c+2m)x^(2)-2(1+2m)xy+(4-c)y^(2)=0` As the lines represented by this pair are perpendicular to each other , we must have Coeffcient of `x^(2)+`Coefficient of `y^(2)=0` (1) Hence , `3c+2m+4-c=0` or `-2=m+c` Comparing this result with `y=mx+c` , we can see that `y=mx+c` (2) passes though (1,-2). |
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| 32. |
If the sum of the slopes of the lines given by `x^2-2c x y-7y^2=0`is four times their product, then the value of `c`is_____ |
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Answer» Given equation of pair of lines is `x^(2)-2cxy-7y^(2)=0` . Let the slopes of the component lines be `m_(1)andm_(2)`. It is given that `m_(1)+m_(2)=4m_(1)m_(2)` `rArr-(2c)/(7)=(-4)/(7)` `:.c=2` |
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| 33. |
The product of perpendiculars let fall from the point `(x_(1),y_(1))` upon the lines represented by `ax^(2)+2hxy+by^(2)`, isA. `(|ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)|)/(sqrt((a-b)^(2)+4h^(2)))`B. `(|ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)|)/(sqrt((a-b)^(2)+h^(2)))`C. `(|ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)|)/(sqrt((a+b)^(2)+4h^(2)))`D. `(|ax_(1)^(2)-2hxy_(1)y_(1)+by_(1)^(2)|)/(sqrt((a-b)^(2)+4h^(2)))` |
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Answer» Correct Answer - A Let `y-x_(1)x=0` and `y=m_(2)x=0` be the lines represented by the equation `ax^(2)+2hxy+by^(2)=0`. Then, `m_(1)+m_(2)=-(2h)/(b) and m_(1)m_(2)=(a)/(b)" (i)"` Let `p_(1) and p_(2)` be the lengths of perpendiculars from `(x_(1),y_(1))` on lines `y-m_(1)x=0` and `y-m_(2)x=0`. Then, `p_(1)=(y_(1)-m_(1)x_(1))/(sqrt(1+m_(1)^(2)))andp_(2)=(y_(1)-m_(2)x_(1))/(sqrt(1+x_(2)^(2)))` `therefore" "p_(1)p_(2)=((y_(1)-m_(1)x_(1))/(sqrt(1+x_(1)^(2))))((y_(1)-m_(2)x_(1))/(sqrt(1+m_(2)^(2))))=((y_(1)-m_(1)x_(1))(y_(1)-m_(2)x_(1)))/(sqrt((1+m_(1)^(2))(1+m_(2))))` `rArr" "p_(1)p_(2)=(y_(1)^(2)-x_(1)y_(1)(m_(1)+m_(2))+m_(1)m_(2)x_(1)^(2))/(sqrt(1+(m_(1)^(2)+m_(2)^(2))+m_(1)^(2)m_(2)^(2)))` `rArr" "p_(1)p_(2)=(y_(1)^(2)-x_(1)y_(1)(m_(1)+m_(2))+m_(1)m_(2)x_(1)^(2))/(sqrt(1+(m_(1)+m_(2))^(2)-2m_(1)m_(2)+(m_(1)m_(2))^(2)))` `rArr" "p_(1)p_(2)=(y_(1)^(2)-x_(1)y_(1)(-(2h)/(b))+(a)/(b)x_(1)^(2))/(sqrt(1+(4h^(2))/(b^(2))-(2a)/(b)+(a^(2))/(b^(2))))" [Using (i)]"` `rArr" "p_(1)p_(2)=(ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2))/(sqrt((a-b)^(2)+4h^(2)))` |
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| 34. |
One bisector of the angle between the lines given by `a(x-1)^(2)+2h(x-1)y+by^(2)=0` is `2x+y-2=0`. The equation of the other bisector isA. `x-2y+1=0`B. `x-2y-2=0`C. `x-2y-1=0`D. none of these |
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Answer» Correct Answer - C We have `a(x-1)^(2)+2h(x-1)(y-0)+b(y-0)^(2)=0` Clearly, it represents a pair of straight lines passing through (1, 0). Since the bisectors of the angles between two lines are always perpendicular. So, the other bisector passes through (1, 0) and is perpendicular to `2x+y-2=0`. Hence, is equation is `y-0=(1)/(2)(x-1)rArrx-2y-1=0`. |
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| 35. |
If the pair of straight lines `a x^2+2h x y+b y^2=0`is rotated about the origin through `90^0`, then find the equations in the new position. |
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Answer» Given equation of pair of straight lines is `ax^(2)+2hxy+by^(2)=0`. Let the component lines be `y=m_(1)xandy=m_(2)x`. `:.m_(1)+m_(2)=(-2h)/(b)andm_(1)m_(2)=(a)/(b)` Now , equations of perpendicular to above component lines are `y=-(1)/(m_(1))xandy=-(1)/(m_(2))x` or `m_(1)y+x=0andm_(2)y+x=0` Combined equation of above lines is `(m_(1)y+x)(m_(2)y+x)=0` or `m_(1)m_(2)y^(2)+xy(m_(1)+m_(2))+x^(2)=0` or `(a)/(b)y^(2)+xy((-2h)/(b))+x^(2)=0` or `bx^(2)-2hxy+ay^(2)=0` Alternative method : We can write the given equation of pair of straight lines as : `b((y)/(x))^(2)+2h((y)/(x))+a=0` The roots of this equations are `m_(1)andm_(2)` which are slopes of component lines . Now , we require equation whose roots are `-(1)/(m_(1))and-(1)/(m_(2))`. So, in above equation , replacing `(y)/(x)by-(x)/(y)`,we get `b(-(x)/(y))^(2)+2h(-(x)/(y))+a=0` or `bx^(2)-2hxy+ay^(2)=0` which is the required equation of pair of straight lines. |
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| 36. |
Prove that the product of the perpendiculars from `(alpha,beta)`to the pair of lines `a x^2+2h x y+b y^2=0`is`(aalpha^2-2halphabeta+bbeta^2)/(sqrt((a-b)^2+4h^2))` |
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Answer» Let the lines be `y=m_(1)xandy=m_(2)x` or `m_(1)x-y=0andm_(2)x-y=0` were `m_(1)+m_(2)=-(2h)/(b),m_(1)m_(2)=(a)/(b)` Now , if the lengths of perpendicular from (0,0) on these lines are `d_(1)andd_(2)` respectively , then `d_(1)d_(2)=(|m_(1)alpha-beta|)/(sqrt(m_(1)^(2)+1))(|m_(2)alpha-beta|)/(sqrt(m_(2)^(2)+1))` `=(|m_(1)m_(2)alpha^(2)-(m_(1)+m_(2))alpha beta+beta^(2)|)/(sqrt(m_(1)^(2)m_(2)^(2)+m_(1)^(2)+m_(2)^(2)+1))` `=(|m_(1)m_(2)alpha^(2)-(m_(1)+m_(2))alpha beta+beta^(2)|)/(sqrt(m_(1)^(2)m_(2)^(2)+(m_(1)+m_(2))^(2)-2m_(1)m_(2)+1))` `(|(a)/(b)alpha^(2)+(2h)/(b)alpha beta+ beta^(2)|)/(sqrt(a^(2)/(b^(2))+(4h^(2))/(b^(2))-2(a)/(b)+1))` `=(|aalpha+2halpha beta+b beta^(2)|)/(sqrt(a^(2)+4h^(2)-2ab+b^(2)))` `=(|aalpha+2halpha beta+b beta^(2)|)/(sqrt((a-b)^(2)+4h^(2)))` |
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| 37. |
One of the bisector of the angle between the lines `a(x-1)^2 + 2h(x-1)(y-2) + b (y-2)^2` = 0 is `x + 2y - 5 = 0`. Then other bisector is(A) `2x-y=0`(B) `2x+y=0`(C) `2x+y-4=0`(D) `x-2y+3=0` |
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Answer» Correct Answer - 1.5 Shifting the origin to (1,2) , the equation of given pair of lines becomes `ax^(2)+2hxy+by^(2)=0` whose one angle bisector is `x+2y=0` (given) and other is `2x-y=0`. `:.` Combined equation of angle bisector is `(x+2y)(2x-y)=0` On comparing with `(x^(2)-y^(2))/(a-b)=(xy)/(h)`, we get `a-b=-3and h=2` `:. (b-a)/(h)=(3)/(2)` |
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| 38. |
If `x^2+2h x y+y^2=0`represents the equation of the straight lines through the origin whichmake an angle `alpha`with the straight line `y+x=0``s e c2alpha=h``cosalpha``=sqrt(((1+h))/((2h)))``2sinalpha``=sqrt(((1+h))/h)`(d) `cotalpha``=sqrt(((1+h))/((h-1)))`A. `sec2alpha=h`B. `cosalpha=sqrt((1+h)//(2h))`C. `2sinalpha=sqrt((1+h)//h)`D. `cot alpha=sqrt((h+1)//(h-1))` |
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Answer» Correct Answer - 1,2,4 Let the equation of the lines given by `x^(2)+2hxy+y^(2)=0` be `y=m_(1)xandy=m_(2)x`. Since these make an angle `alpha` with `y+x=0` whose slope is -1, we have `(m_(1)+1)/(1-m_(1))=tanalpha=(-1-m_(2))/(1-m_(2))` or `m_(1)+m_(2)=((tanalpha-1)^(2)+(tanalpha+1)^(2))/(tan^(2)alpha-1)` `=(-sec^(2)alphaxxcos^(2)alpha)/(cos2alpha)` `:.-2sec2alpha=-2h` or `sec2alpha=h` or `cos2alpha=(1)/(h)or2cos^(2)alpha-1=(1)/(h)` or `cosalpha=sqrt((1+h)/(2h))andcotalpha=sqrt((h+1)/(h-1))` |
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| 39. |
Prove that the straight lines joining the origin to the point ofintersection of the straight line `h x+k y=2h k`and the curve `(x-k)^2+(y-h)^2=c^2`are perpendicular to each other if `h^2+k^2=c^2dot`A. `h^(2)+k^(2)=c^(2)`B. `h^(2)+k^(2)=2c^(2)`C. `h^(2)-k^(2)=c^(2)`D. none of these |
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Answer» Correct Answer - A We have, `kx+hy=2hx" …(i)"` and, `x^(2)+y^(2)-2(hx+ky)+h^(2)+k^(2)-c^(2)=0" (ii)"` The combined equation of the straight lines joining the origin to the points of intersection of (i) and (ii) is `x^(2)+y^(2)-2(hx+ky)((kx+hy)/(2hk))+(h^(2)+k^(2)-c^(2))((kx+hy)/(2hk))^(2)=0` `rArr" "(h^(2)y^(2)+k^(2)x^(2))(h^(2)+k^(2)-c^(2))-2hk(h^(2)+k^(2)+c^(2))xy=0` Lines given by this equation are at right angle. `therefore" "(k^(2)+h^(2))(h^(2)+k^(2)-c^(2))=0rArrh^(2)+k^(2)=c^(2)` |
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| 40. |
The angle between the straight lines joining the origin to the point of intersection of `3x^(2)+5xy-3y^(2)+2x+3y=0` and `3x-2y=1`, isA. `(pi)/(3)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(2)` |
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Answer» Correct Answer - D The combined equation of the straight lines joining the origin to the points of intersection of `3x^(2)+5xy-3y^(2)+2x+3y=0 and 3x-2y=1` is `3x^(2)+5xy-3y^(2)+(2x+3y)(3x-2y)=0` `rArr" "9x^(2)+10xy-9y^(2)=0` Clearly, it represents a pair of perpendicular straight lines. |
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| 41. |
The equation `a^2x^2+2h(a+b)x y+b^2y^2=0`and `a x^2+2h x y+b y^2=0`representtwo pairs of perpendicular straight linestwo pairs of parallel straight linestwo pairs of straight lines which are equally inclined to each othernone of theseA. two pair of perpendicular straight linesB. two pairs of parallel straight linesC. two pairs of straight lines which are equally inclined to each otherD. None of these |
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Answer» Correct Answer - 3 The given equations are `a^(2)x^(2)+2h(a+b)xy+b^(2)y^(2)=0` (1) and `ax^(2)+2hxy+by^(2)=0` (2) The equation of the bisectors of the angles between the lines represented by (1) is `(x^(2)-y^(2))/(a^(2)-b^(2))=(xy)/(h(a+b))` or `(x^(2)-y^(2))/(a-b)=(xy)/(h)` which is the same as the equation of bisectors of the angles between the line pair (2). thus , two line pairs are equally inclined to each other. |
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| 42. |
Prove that the straight lines joining the origin to the points of intersection of the straight line `hx+ky=2hk` and the curve `(x-k)^(2)+(y-h)^(2)=c^(2)` are at right angle if `h^(2)+k^(2)=c^(2)`. |
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Answer» From the equation of line , we have `1=(hx+ky)/(2hx)` (1) Equation of the curve is `x^(2)+y^(2)-2kx-2hy+h^(2)+k^(2)-c^(2)=0` Making above equation homogeneous with the help of (1) , we get `x^(2)+y^(2)-2(kx+hy)((hx+ky)/(2hx))+(h^(2)+k^(2)-c^(2))((hx+ky)/(2hk))^(2)=0` This is combined equation of the pair of lines joining the origin to the points of intersection of the given line and the curve . the componenet lines are perpendicular if sum of coefficient of `x^(2)` of and coefficient of `y^(2)` is zero . ` :. (h^(2)+k^(2))(h^(2)+k^(2)-c^(2))=0` `rArr h^(2)+k^(2)=c^(2)` `(ash^(2)+k^(2)ne0)` |
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| 43. |
If the lines joining the origin to the points of intersection of the line `y=mx+2` and the curve `x^(2)+y^(2)=1` are at right-angles, thenA. `m^(2)=1`B. `m^(2)=2`C. `m^(2)=7`D. `2m^(2)=1` |
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Answer» Correct Answer - C The joint equation of the lines joining the origin and the point of intersection of the line `y=mx+2` and the curve `x^(2)+y^(2)=1` is `x^(2)+y^(2)=((y-mx)/(2))^(2)rArrx^(2)(4-m^(2))+2mxy+3y^(2)=0` The lines given by the above equation are at right angles. `therefore" "4-m^(2)+3=0rArrm^(2)=7` |
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| 44. |
Find the angle between the straight lines joining the origin to thepoint of intersection of `3x^2+5x y-3y^2+2x+3y=0`and `3x-2y=1` |
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Answer» Correct Answer - `pi//2` The equation of the line is `3x-2y=1` (1) and the equation of the curve is `3x^(2)+5xy-3y^(2)+2x+3y=0` (2) Making (2) homogenous with the help of (1), we get `3x^(2)+5xy-3y^(2)+(2x+3y)(3x-2y)=0` or `9x^(2)+10xy-9y^(2)=0` (3) which is the equation of the lines joining the origin to the point of intersection of (1)and (2). Here ,` a=9,b=-9`. Since a+b=0, the two lines given by (3) are at right angles . |
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| 45. |
If the equation `x^(4)+bx^(3)y+cx^(2)y^(2)+dxy^(3)+ey^(4)=0` represent two pairs of perpendicular lines, thenA. `b+d=1 and e=-1`B. `b+d=0 and e=-1`C. `b+d=0 and e=1`D. none of these |
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Answer» Correct Answer - B Let `x^(2)+alpha xy-y^(2)=0 and x^(2)+beta xy-y^(2)=0` be two pairs of perpendicular lines given by the equation `x^(4)+bx^(3)y+cx^(2)y^(2)+dxy^(3)+ey^(4)=0`. Then, `x^(4)+bx^(3)y+cx^(2)y^(2)+dxy^(3)+ey^(4)=(x^(2)+alpha xy-y^(2))(x^(2)+beta xy-y^(2))` On equating the coefficients of like terms, we get `e=-1, b=beta+alpha, c=alpha beta and d=-beta-alpha` `rArr" "b+d=-, e=-1` |
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| 46. |
Equation of pair of straight lines drawn through (1, 1) and perpendicular to the pair of lines `3x^(2)-7xy+2y^(2)=0`, isA. `2x^(2)+7xy-11x+6=0`B. `2(x-1)^(2)+7(x-1)(y-1)-3y^(2)=0`C. `2(x-1)^(2)+7(x-1)(y-1)-3(y-1)^(2)=0`D. none of these |
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Answer» Correct Answer - D The joint equation of pair of straight lines passing through the origin and perpendicular to the pair of lines `3x^(2)-7xy+2y^(2)=0` is `2x^(2)+7xy+3y^(2)=0`. Shifting the origin at (1, 1), we obtain `2(x-1)^(2)+7(x-1)(y-1)+3(y-1)^(2)=0` as the joint equation of pair of lines drawn through (1, 1) and perpendicular to the pair of lines `3x^(2)-7xy+2y^(2)=0`. |
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| 47. |
Find the equation of the bisectors of the angles between the linesjoining the origin to the point of intersection of the straight line `x-y=2`with the curve `5x^2+11 x y=8y^2+8x-4y+12=0` |
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Answer» Correct Answer - `x^(2)+30xy-y^(2)=0` The equation of lines joining the origin to the points of intersection of the given line and curve is `5x^(2)+11xy-8y^(2)+(8x-4y)((x-y)/(2))+12((x-y)/(2))^(2)=0` or `12x^(2)-xy-3y^(2)=0` The equation of bisectors is `(x^(2)-y^(2))/(12-(-3))=(xy)/(-1//2)` or ` x^(2)+30xy-y^(2)=0` |
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| 48. |
The lines joining the origin to the point of intersection of `x^2+y^2+2gx+c=0 and x^2+y^2+2fy-c=0` are at right angles ifA. `g^(2)+f^(2)=c`B. `g^(2)-f^(2)=c`C. `g^(2)-f^(2)=2c`D. `g^(2)+f^(2)=c^(2)` |
| Answer» Correct Answer - C | |
| 49. |
The lines joining the origin to the point of intersection of The linesjoining the origin to the point of intersection of `3x^2+m x y=4x+1=0`and `2x+y-1=0`are at right angles. Then which of the following is not a possible valueof `m ?``-4`(b) 4(c) 7 (d)3A. -4B. 4C. 7D. 3 |
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Answer» Correct Answer - 1,2,3,4 The equation of the lines joining the origin to the points of intersection of the given lines is `3x^(2)+mxy-4x(2x+y)+(2x+y)^(2)=0` (by homogenization) or `x^(2)=mxy-y^(2)=0` which are perpendicular for all values of m. |
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| 50. |
If the equation `ax^2+by^2+cx+cy=0` represents a pair of straight lines , thenA. not realB. coincidentC. mutually perpendiculatD. strictly parallel |
| Answer» Correct Answer - C | |