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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.
| 101. |
Joint equation of two lines, through the origin, such that one of them is parallel and the other perpendicular to line `2x+3y+c=0` isA. `6x^(2)-5xy-6y^(2)=0`B. `6x^(2)-5xy+6y^(2)=0`C. `6x^(2)+5xy-6y^(2)=0`D. `6x^(2)+5xy+6y^(2)=0` |
| Answer» Correct Answer - C | |
| 102. |
Joint equation of two lines through the origin, such that one is parallel to line `x+2y=5` and the other perpendicular to line `2x-y+3=0` isA. `x^(2)-4x-4y^(2)=0`B. `x^(2)-4xy+4y^(2)=0`C. `x^(2)+4xy-4y^(2)=0`D. `x^(2)+4xy+4y^(2)=0` |
| Answer» Correct Answer - D | |
| 103. |
Joint equation of two lines, through the origin, each making an angle of `30^(@)` with the Y-axis isA. `x^(2)-3y^(2)=0`B. `3x^(2)-y^(2)=0`C. `2x^(2)-3y^(2)=0`D. `x^(2)+3y^(2)=1` |
| Answer» Correct Answer - B | |
| 104. |
Joint equation of two lines, through the origin, each making an angle of `30^(@)` with the X-axis isA. `x^(2)-3y^(2)=0`B. `3x^(2)-y^(2)=0`C. `2x^(2)-3y^(2)=0`D. `3x^(2)-y^(2)=1` |
| Answer» Correct Answer - A | |
| 105. |
Joint equation of two lines through (2,-3) perpendicular to two lines `3x^(2)+xy-2y^(2)=0` isA. `2x^(2)+xy-3y^(2)-5x-20y-25=0`B. `-2x^(2)-xy+3y^(2)-5x-20y-25=0`C. `3x^(2)+xy-2y^(2)-5x-20y-25=0`D. `2x^(2)+xy-3y^(2)+5x+20y-25=0` |
| Answer» Correct Answer - A | |
| 106. |
Joint equation of two lines both parallel to Y-axis and each at a distance of 3 units from it isA. `x^(2)-9=0`B. `y^(2)-9=0`C. `x^(2)-y^(2)=9`D. `y^(2)+9=0` |
| Answer» Correct Answer - A | |
| 107. |
Find the distance between the pair of parallel lines `x^2+4x y+4y^2+3x+6y-4=0` |
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Answer» Given lines are `(x+2y)^(2)+3(x+2y)-4=0` `:. x+2y=(-3+-sqrt(9+16))/(2)=(-3+-5)/(2)=-4,1` therefore , the lines are `x+2y+4=0` `x+2y-1=0` Required distance `=(|4-(-1)|)/(sqrt(1+4))` `=(5)/(sqrt(5))=sqrt(5)` |
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| 108. |
Two lines jointly given by the equation `xy-2y+y-2=0` areA. `||` to coordinate axes separately and `_|_` to each otherB. `_|_` to coordinate axes separately and `_|_` to each otherC. `||` as well as `_|_` to coordinatesD. `||` and `_|_` to coordinates axes, and `_|_` to each other |
| Answer» Correct Answer - D | |
| 109. |
Mesure of angle between the two lines `x^(2)(cos^(2)theta-1)-xy sin^(2) theta+y^(2)sin^(2) theta=1` isA. `(pi)/3`B. `(pi)/4`C. `(2pi)/3`D. `(pi)/2` |
| Answer» Correct Answer - D | |
| 110. |
Joint equation of the two lines `x+y=1` and `x-y=4` isA. `x^(2)-y^(2)=-4`B. `x^(2)-y^(2)=4`C. `(x+y-1)(x-y-4)=0`D. `(x+y+1)(x-y+4)=0` |
| Answer» Correct Answer - C | |
| 111. |
Given pair of lines `2x^(2) +5xy +2y^(2) +4x +5y +a = 0` and the line `L: bx +y +5 = 0`. ThenA. `a = 2`B. `a =- 2`C. There exists no circle which touches the pair of lines and the line L if `b = 5`.D. There exists no circle which touches the pair of lines and the line L if `b =- 5` |
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Answer» Correct Answer - A::C `2x^(2) +5xy +2y^(2) +4x 5y +a =0` represents the pair of lines if `|{:(2,5//2,2),(5//2,2,5//2),(2,5//2,a):}| =0` `rArr a = 2` So pair of lines is `2x^(2)+5xy +2y^(2) +4x +5y +2 =0` or `x +2y +1 = 0, 2x +y +2 =0` These lines are concurrent with `bx +y +5 =0` If `|{:(1,2,1),(2,1,2),(b,1,5):}| =0` `rArr b =5` Which lines are concurrent, no circle can be drawn touching all three lines. |
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| 112. |
`9x^(2) +2hxy +4y^(2) +6x +2fy - 3 = 0` represents two parallel lines. ThenA. `h = 6, f = 2`B. `h =- 6, f = 2`C. `h = 6, f =- 2`D. `h =- 6, f =- 2` |
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Answer» Correct Answer - A::D Since the given equation represents a pair of parallel lines, we have `h^(2) = ab rArr h = +-6` Condition for pair of lines `|{:(9,h,3),(h,4,f),(3,f,-3):}| =0` `rArr 9f^(2) -6hf +36 = 0` For `h = 6, f^(2) -4f +4 =0` or `f =2` For `h =- 6, f^(2)+4f +4 =0` or `f =- 2` |
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| 113. |
If the slope of one of the lines given by `ax^(2)-6xy+y^(2)=0` is twice the other, then a =A. 1B. 2C. 4D. 8 |
| Answer» Correct Answer - D | |
| 114. |
Find the lines whose combined equation is `6x^2+5x y-4y^2+7x+13 y-3=0` |
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Answer» The given equation of pair of straight lines is `6x^(2)+5xy-4y^(2)+7x+13y-3=0` or `6x^(2)+(5y+7)x-(4y^(2)-13y+3)=0` Solving it as a quadratic in x,we get `x=((5y+7)+-sqrt((5y+7)^(2)+24(4y^(2)-13y+3)))/(12)` `=(-(5y+7)+-sqrt(121y^(2)-242y+121))/(12)` `=(-(5y+7)+-11(y-1))/(12)` `=(6y-18)/(12),(16y+4)/(12)or(y-3)/(2),(-4y+1)/(3)` The two straight lines are `2x-y++3=0and3x+4y-1=0`. |
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| 115. |
If `theta`is the angle between the lines given by the equation `6x^2+5x y-4y^2+7x+13 y-3=0`, then find the equation of the line passing through the point ofintersection of these lines and making an angle `theta`with the positive x-axis.A. `2x+11y+13=0`B. `11x-2y+13=0`C. `2x-11y+2=0`D. `11x+2y-11=0` |
| Answer» Correct Answer - B | |
| 116. |
The diagonal of the rectangle formed by the lines `x^2-7x +6= 0` and `y^2-14y +40= 0` isA. `5x+6y=0`B. `5x-6y=0`C. `6x-5y+14=0`D. `6x-5y-14=0` |
| Answer» Correct Answer - C | |
| 117. |
The equation of the diagonal of the square formed by the pairs of lines `xy +4x - 3y - 12 = 0` and `xy - 3x +4 y - 12 = 0` isA. `x - y = 0`B. `x +y +1 = 0`C. `x +y = 0`D. `x - y +1 = 0` |
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Answer» Correct Answer - A::B `xy +4x -3y -12 =0` `rArr (x-3) (y+4) =0` `xy -3x +4y -12 =0` `rArr (x-4) (y-3) =0` The vertices are `A =(-4,-4), B =(-4,3), C =(3,3), D =(3,-4)` Diagonal AC is `x =y` and diagonal BD is `x +y +1 = 0`. |
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| 118. |
The equation `x^3-yx^2+x-y= 0` representsA. a hyperboalB. an ellipseC. a pair of straight linesD. a rectangular hyperbola |
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Answer» Correct Answer - C We, have `y^(2)-x^(2)+2x-1=0` `rArr" "y^(2)-(x-1)^(2)=0` `rArr" "(y+x-1)(y-x+1)=0rArry+x-1=0, y-x+1=0` Hence, the given equation represents a pair of straight lines. `ul("ALITER")` Comparing the given equation with `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0` we get `a=-1,b=1,g=1, c=-1, f=h=0` `therefore" "abc+2fgh-af^(2)-bg^(2)-ch^(2)=1+0-0-1-0=0` Hence, the given equation represents a pair of straight lines. |
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| 119. |
If the equation `4y^(3) - 8a^(2)yx^(2) - 3ay^(2)x +8x^(3) = 0` represents three straight lines, two of them are perpendicular, then sum of all possible values of a is equal toA. `(3)/(8)`B. `(-3)/(4)`C. `(1)/(4)`D. `-2` |
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Answer» Correct Answer - B We have `4y^(3) -8a^(2)yx^(2) - 3ay^(2)x +8x^(3) =0` `rArr 4 ((y)/(x))^(3) -3a ((y)/(x))^(2) -8a^(2) ((y)/(x)) +8 =0` has roots `m_(1),m_(2),m_(3)` `:. m_(1)m_(2)m_(3) =- 2` Given `m_(1)m_(2) =-1` `:. m_(3)=2` `:. 4(2)^(3)-3a(2)^(2) -8a^(2)(2) +8 =0` `rArr 4a^(2) +3a - 10 =0` `:.` Sum of possible values of roots `=(-3)/(4)` |
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| 120. |
The orthocenter of the triangle formed by the lines `xy=0 and x+y=1` isA. `(1//2,1//2)`B. `(1//3,1//3)`C. `(0,0)`D. `(1//4,1//4)` |
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Answer» Correct Answer - 3 The lines by which triangle is formed are `x=0,y=0and x+y=1`. Clearly , it is a right triangle . We know that in a right angled triangle , the orthocenter coincides with the vertex at which right angles is formed . Therefore , the orthocenter is (0,0). |
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| 121. |
Statement1 : If `-h2=a+b ,`then one line of the pair of lines `a x^2+2h x y+b y^2=0`bisects the angle between the coordinate axes in the positive quadrant.Statement2 : If `a x+y(2h+a)=0`is a factor of `a x^2+2h x y+b y^2=0,`then `b+2h+a=0`Both the statements are true but statement 2 is the correct explanationof statement 1.Both the statements are true but statement 2 is not the correctexplanation of statement 1.Statement 1 is true and statement 2 is false.Statement 1 is false and statement 2 is true.A. Both the statements are true but statement 2 is the correct explanation of statement 1.B. Both the statements are true but statement 2 is not the correct explanation of statement 1.C. Statement 1 is true and statement 2 is false.D. Statement 1 is false and statement 2 is true. |
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Answer» Correct Answer - 2 Put `2h=-(a+b)`in`ax^(2)+2hxy+by^(2)=0`. Then , `ax^(2)-(a+b)xy+by^(2)=0` or `(x-y)(ax-by)=0` Therefore , one of the lines bisects the angle between the coordinates axes in the positive quadrant. Also , putting `b=-2h-a` in `ax-by` ,we have `ax-by=ax-(-2h-a)y=ax+(2h+a)y` Hence , `ax+(2h+a)y` is a factor of `ax^(2)+2hxy+by^(2)` . However, statement 2 is not the correct explanation of statement 1. |
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| 122. |
Consider the equation of a pair of straight lines as `lambdax^(2)-10xy+12y^(2)+5x-16y-3=0`. The point of intersection of lines is `(alpha, beta)`. Then the value of `alpha beta` isA. 35B. 45C. 20D. 15 |
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Answer» Correct Answer - 1 `2x^(2)=10xy+12y^(2)+5x-16y-3=0` Consider the homogeneous part `2x^(2)-10xy+12y^(2)=(x-2y)(2x-6)` `2x^(2)-10xy+12y^(2)+5x-16y-3` `-=(2x-6y+A)(x-2y+B)` Comparing coefficients , we get `A=-1,B=3` Hence , the lines are `2x-6y-1=0and x-2y+3=0` Solving , we get the intersection points as `(-10,-7//2)`. Therefore , Product `=35` |
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| 123. |
The straight lines represented by `(y-mx)^(2)=a^(2)(1+m^(2))and(y-nx)^(2)=a^(2)(1+n^(2))`form aA. rectangleB. rhombusC. trepeziumD. None of these |
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Answer» Correct Answer - 2 The straight lines represented by `(y-mx)^(2)=a^(2)(1+m^(2))` are `y-mx=+-asqrt(1+m^(2))` i.e., `y-mx=asqrt(1+m^(2))` (1) and `y-mx=-asqrt(1+m^(2))` (2) Similarly , the straight lines representd by `(y-nx)^(2)=a^(2)(1+n^(2))` are `y-nx=asqrt(1+n^(2))` (3) and ` y-nx=-asqrt(1+n^(2))` (4) Since the lines (1) and (2) are parallel , the distance between thems is `|(asqrt(1+m^(2))+asqrt(1+m^(2)))/(sqrt(1+m^(2)))| =|2a|` Similarly , the lines (3) and (4) are parallel lines and the distance between them is `|2a|` . Since the distances between parallel lines are the same , the four lines form a rhombus. |
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| 124. |
The equation `x^2 - 3xy+ lambday^2 + 3x - 5y + 2 = 0` where `lambda` is a real number, represents a pair of straight lines. If `theta` is the angle between the lines, then `cosec^2theta =`A. 2B. 0C. 3D. 1 |
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Answer» Correct Answer - B We have, `x^(2)-3xy+lambda y^(2)+3x-5y+2=0` Here, `a=1, b=lambda,c=2, h=-3//2,g=3//2 and f=-5//2` We have, `tan(tan^(-1)3)=(2sqrt((9)/(4)-lambda))/(1+lambda)` `rArr" "3=(sqrt(9-4lambda))/(a+lambda)` `rArr" "9(lambda+1)^(2)=9-4lambda` `rArr" "9lambda^(2)+22lambda=0rArr lambda=0,-(22)/(9)rArr lambda=0" "[because lambda ge 0]` |
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| 125. |
The equation `x^(3)+x^(2)y-xy^(2)=y^(3)` representsA. three real straight linesB. lines in which two of them are perpendicular to each otherC. lines in which two of them are coincidentD. None of these |
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Answer» Correct Answer - 1 , 2 , 3 The equation is `x^(2)(x+y)-y^(2)(x+y)=0` or `(x+y)^(2)(x-y)=0` It represents the lines `x+y=0,x+y=0,x-y=0`. |
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| 126. |
The two lines represented by `3ax^(2)+5xt+(a^(2)-2)y^(2)=0` are perpendicular to each other forA. two values of aB. aC. for one value of aD. for no value of a |
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Answer» Correct Answer - 1 `3a+a^(2)-2=0` or `a^(2)+3a-2=0` or `a=(-3+-sqrt(9+8))/(2)=(-3+-sqrt(17))/(2)` Thererfore , there are two values of a . |
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| 127. |
The equation `x-y=4 and x^2+4xy+y^2=0` represent the sides ofA. an equilateral triangleB. a right - angled triangleC. an isosceles triangleD. None of these |
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Answer» Correct Answer - 1 Acute angle between the linens `x^(2)+4xy+y^(2)=0` is `tan^(-1){(2sqrt(4-1))//(1+1)}=tan^(-1)sqrt(3)=pi//3`. the angle bisectors of `x^(2)+4xy+y^(2)=0`are given by `(x^(2)-y^(2))/(1-1)=(xy)/(2)` or `x^(2)-y^(2)=0orx=+-y` As `x+y=0` is perpendicular to `x-y=4` , the given triangle is isosceles with vertical angle equal to `pi//3`. Hence , it is an equilateral triangle. |
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| 128. |
The equation `x^2 - 3xy+ lambday^2 + 3x - 5y + 2 = 0` where `lambda` is a real number, represents a pair of straight lines. If `theta` is the angle between the lines, then `cosec^2theta =`A. 9B. 10C. 15D. 26 |
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Answer» Correct Answer - B The equation `x^(2)-3xy+lambda y^(2)+3x-5y+2=0` will represent a pair of straight lines, if `2lambda+2xx-(5)/(2)xx(3)/(2)xx-(3)/(2)-(-(5)/(2))^(2)-lambda((3)/(2))^(2)-2(-(3)/(2))^(2)=0` `rArr" "8 lambda+45-25-9lambda-18=0rArr lambda =2` Now, `tantheta=(2sqrt(h^(2)-ab))/(a+b)` `rArr" "tan theta=(2sqrt((9)/(4)-2))/(1+2)=(1)/(3)rArr" cosec"^(2)theta=1+9=10` |
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| 129. |
Consider the equation of a pair of straight lines as `lambdax^(2)-10xy+12y^(2)+5x-16y-3=0`. The angles between the lines is `theta` . Then the value of `tan theta` isA. `1//5`B. `2//9`C. `1//7`D. `3//4` |
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Answer» Correct Answer - 3 `tantheta=(2sqrt(h^(2)-ab))/(a+b)=(2sqrt(25-24))/(14)=(1)/(7)` |
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| 130. |
The distance between the two lines represented by the sides ofan equilateral trianglea right-angled trianglean isosceles trianglenone of theseA. `8//5`B. `6//5`C. `11//5`D. None of these |
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Answer» Correct Answer - 1 `9x^(2)-24xy+16y^(2)-12x+16y-12=0` or `(3x-4y+2)(3x-4y-6)=0` Hence , the distance between the lines is `(|6-(-2)|)/(5)=(8)/(5)`. |
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| 131. |
If one of the lines represented by the equation `lambda y^(2)+(1-lambda^(2))xy-lambdax^(2)=0` is bisector of the angle between the lines `xy=0`, then, `lambda=`A. `1,-1`B. `2,-(1)/(2)`C. `1,2`D. `-1,-(1)/(2)` |
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Answer» Correct Answer - A We have, `xy=0 rArr x = 0 or y =0 ` Thus, xy = 0 represent the coordinate axes. The equation of the angle bisectors between the coordinate axes are y = x and `y=-x`. It is given that one of the lines represented by the equaiton `lambday^(2)+(1-lambda^(2))xy- lambda x^(2)=0" ....(i)"` is `y=x or , y=-x` `therefore" "lambda+1-lambda^(2)-lambda=0" "["Putting y = x on (i)"]` or, `lambda-(1-lambda^(2))-lambda=0" "["Putting "y=-x " in (i)"]` `rArr" "1-lambda^(2)=0 rArr lambda = pm 1` |
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| 132. |
The straight lines represented by `(y-m x)^2=a^2(1+m^2)`and `(y-n x)^2=a^2(1+n^2)`from arectangle(b) rhombustrapezium(d) none of theseA. ractangleB. trapeziumC. rhombusD. none of these |
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Answer» Correct Answer - C The equations of the straight lines represebted by the two equations are `{:(y=mx+asqrt(1+m^(2))" ...(i) "y=mx-asqrt(1+m^(2))" ...(ii)"),(y-nx+asqrt(1+n^(2))" ...(iii) "y-nx-asqrt(1+n^(2))" ...(iv)"):}` Clearly, we have two sets of parallel straight lines. So, they form a parallelogram. Also, the distance between (i) and (ii) is same as the distance between (iii) and (iv) each equal to `2|a|`. So, the given lines form a rhombus. |
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| 133. |
If two lines represented by `x^4+x^3y+c x^2y^2-x y^3+y^4=-`bisector of the angle between the other two, then the value of `c`is`0`(b) `-1`(c) 1 (d)`-6` |
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Answer» Correct Answer - 4 Since the product of the slope of the four lines represented by the given equation is 1 and a pair of lines represents the bisectors of the angles between the other two , the product of the slopes of each pairs is -1. So , let the equation of one pairs be `ax^(2)+2hxy-ay^(2)=0`. then the equation of its bisectors is `(x^(2)-y^(2))/(2a)=(xy)/(h)` By hypothesis , `x^(4)_x^(3)y+cx^(2)y^(2)-xy^(3)+y^(4)=(ax^(2)+2hxy-ay^(2))(hx^(2)-2axy-hy^(2))=ah(x^(4)+y^(4))+2(h^(2)-a^(2))(x^(3)y-xy^(3))-6ahx^(2)y^(2)` |
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| 134. |
The straight lines represented by `x^2+m x y-2y^2+3y-1=0`meet at`(-1/3,2/3)`(b) `(-1/3,-2/3)``(1/3,2/3)`(d) none of theseA. `(-1//3,2//3)`B. `(-1//3,-2//3)`C. `(-1//3,-2//3)`D. None of these |
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Answer» Correct Answer - 1, 3 The equation represents a pair of straight lines . Hence, `1xx(-2)(-1)+2((3)/(2))xx0xx(m)/(2)-1xx((3)/(2))^(2)-(-2)xx0^(2)-(-1)xx((m)/(2))^(2)=0` or `m=1,-1` The point of intersection of the pair of lines are obtained by solving `(partialS)/(partial x)-=2x+my=0and (partialS)/(partialy)-=mx-4y+3=0` When `m=1`, the required point is the intersection of `2x+y=0,x-4y+3=0` When `m=-1`, the reqired point is the intersection of `2x-y=0,-x4y+3=0`. |
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| 135. |
The lines represented by `x^(2)+2lambda xy+2y^(2)=0` and the lines represented by `1+lambda_x^(2)-8xy+y^(2)=0` are equally inclined, thenA. `lambda` is any real numberB. `lambda gt 2`C. `lambda = pm2`D. `lambda lt -2` |
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Answer» Correct Answer - C If the lines given by the two equations are equally inclined then they have the same bisectors. Therefore, equations `(x^(2)-y^(2))/(1-2)=(xy)/(lambda) and (x^(2)-y^(2))/(1+lambda-1)=(xy)/(-4)` represent the same pair of lines. `rArr" "lambdax^(2)+xy-lambday^(2)=0 and 4x^(2)+lambdaxy-4y^(2)=0`, represent the same pair of lines `therefore" "(lambda)/(4)=(1)/(lambda)=(-lambda)/(-4)rArr lambda^(2)=4 rArr lambda= pm 2` |
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| 136. |
The straight lines represented by the equation `135 x^2-136 x y+33 y^2=0`are equally inclined to the line`x-2y=7`(b) x+2y=7`x-2y=4`(d) `3x+2y=4`A. `x-2y=7`B. `x+2y=7`C. `x-2y=4`D. `3x+2y=4` |
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Answer» Correct Answer - 2 The given pair of lines is `135x^(2)-136xy+33y^(2)=0` (1) the equation of bisectors of angles between the pairs of lines (1) is `(x^(2)-y^(2))/(a-b)=(xy)/(h)or(x^(2)-y^(2))/(135-33)=(xy)/(-68)` or `2x^(2)+3xy-2y^(2)=0`(2) or `(x+2)(2x-y)=0` One of the bisectors is `x+2y=0` which is parallel to the line `x+2y=7`. Hence , the line `x+2y=7` is equally inclined to the given lines. |
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| 137. |
if `X^2/a+y^2/b+(2xy)/h=0` represent pair of straight lies and slope one line is twice the other line then `ab:h^2`.A. `1:2`B. `2:1`C. `8:9`D. `9:8` |
| Answer» Correct Answer - D | |
| 138. |
Equation `x^(2)y^(2)-9y^(2)+6x^(2)y-54y=0` representsA. a pair of lines and a circleB. a pair of lines and a parabolaC. a set of four lines which form a squareD. a set of four lines along a rectangle |
| Answer» Correct Answer - C | |
| 139. |
The diagonals of a square are along the pair of lines whose equation is `2x^2 - 3xy - 2y^2 = 0`If (2,1) is a vertex of the square, then the vertex of the square adjacent to it may beA. (1,4)B. (1,-2)C. (2,-1)D. (1,2) |
| Answer» Correct Answer - B | |
| 140. |
The value `k`for which `4x^2+8x y+k y^2=9`is the equation of a pair of straight lines is__________ |
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Answer» Correct Answer - 2 The equation `4x^(2)+8xy+ky^(2)-9=0` represents a pair of straight lines if `(4)(k)(-9)-(4)^(2)=0ork=4`. |
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| 141. |
Area of the triangle formed by the lines `y^2-9x y+18 x^2=0a n dy=6`is____ |
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Answer» Correct Answer - 3 We have `y^(2)-9xy+18x^(2)=0` or `y^(2)-6xy-3xy+18x^(2)=0` or `y(y-6x)-3x(y-6x)=0` or `y-3x=0andy-6x=0` The third line is `y=6` . Therefore , the area of the triangle formed by these lines `=(1)/(2)|{:(0,0,1),(1,6,1),(2,6,1):}|` `=(1)/(2)|6-12|` `=3"units"^(2)` |
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| 142. |
The distance between the lines `(x+7y)^2+4sqrt(2)(x+7y)-42=0`is_____________ |
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Answer» Correct Answer - 2 `(x+7y)=4sqrt(2)(x+7y)-42=0` or `(x+7y)^(2)+7sqrt(2)(x+7y)-3sqrt(2)(x+7y)-42=0` or `(x+7y)[x+7y+7sqrt(2)]-3sqrt(2)(x+7y+7sqrt(2))=0` or `(x+7y+7sqrt(2))(x+7y-3sqrt(2))=0` or `x+7y+7sqrt(2)=0andx+7y-3sqrt(2)=0` or `d=|(7sqrt(2)+2sqrt(2))/(sqrt(1+49))|=(10sqrt(2))/(sqrt(50))=2` |
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| 143. |
If the line passing through `P(1,2)` making an angle `45^(@)` with the x-axis in the positive direction meets the pair of lines `x^(2) +4xy +y^(2) = 0` at A and B then `PA. PB =`A. `13//3`B. `13//6`C. `11//6`D. `11//3` |
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Answer» Correct Answer - A Any point at distance r from the point `P(1,2)` on the line making an angle `45^(@)` with the x-axis is `(1+(r )/(sqrt(2)),2+(r )/(sqrt(2)))` Solving this point with the given pair of lines `x^(2)+4xy +y^(2) =0`, we get `3r^(2) +9sqrt(2)r +13 =0` Product of roots `r_(1)r_(2) = 13//3` |
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| 144. |
A line passes through (2,0). Then which of the following is not the slope of the line, for which its intercept between `y=x-1` and `y=-x+1` subtends a right angle at the origin?A. `-(1)/(sqrt(3))`B. `-sqrt(3)`C. `(1)/(sqrt(3))`D. None of these |
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Answer» Correct Answer - B The joint equation of straight line `y = x -1` and `y =- x +1` is `(x-y-1) (x+y-1) =0` `rArr x^(2) -y^(2) -2x +1 =0` (i) Let equation of line passes through (2,0) is `y = m (x-2)` By homogenizing equation (i) with help of line (ii), we get `x^(2)-y^(2)-2x ((mx-y)/(2m)) +((mx-y)/(2m))^(2) =0` This is pair of straight line through origin. Since these lines are perpendicular, Coefficiennt of `x^(2) +` coefficient of `y^(2) =0` `rArr m = +- (1)/(sqrt(3))` |
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| 145. |
The value of `lambda` for which the equation `x^2-y^2 - x - lambda y - 2 = 0` represent a pair of straight line, areA. `-3,1`B. `3,-3`C. `-1,1`D. `3,1` |
| Answer» Correct Answer - B | |
| 146. |
The equation of second degree `x^2+2sqrt2x+2y^2+4x+4sqrt2y+1=0` represents a pair of straight lines.The distance between them isa. 4b. `4/sqrt3`c. 2d. `2sqrt3`A. 4B. `2sqrt3`C. `4sqrt3`D. 2 |
| Answer» Correct Answer - D | |
| 147. |
If the slope of one of the lines given by `ax^(2)-6xy+y^(2)=0` is square of the other, then a =A. `8, -27`B. `-8, 27`C. `1,8`D. `-8,-27` |
| Answer» Correct Answer - A | |
| 148. |
The joint equation of the straight lines `x+y=1` and `x-y=4`, isA. `x^(2)-y^(2)=-4`B. `x^(2)-y^(2)=4`C. `(x+y-1)(x-y-4)=0`D. `(x+y+1)(x-y+4)=0` |
| Answer» Correct Answer - C | |
| 149. |
Distance between the lines represented by `9x^2-6x y+y^2+18 x-6y+8=0`, is`1/(sqrt(10))`2.`2/(sqrt(10))`3.`4/(sqrt(10))`4. `sqrt(10)`5.`2/(sqrt(5))`A. `(1)/(sqrt(10))`B. `(2)/(sqrt(10))`C. `(4)/(sqrt(10))`D. `sqrt(10)` |
| Answer» Correct Answer - B | |
| 150. |
The circumcentre of the triangle formed by the lines, `xy + 2x + 2y + 4 = 0 and x + y + 2 = 0` is-A. `(0,0)`B. `(-2,-2)`C. `(-1,-1)`D. `(-1,-2)` |
| Answer» Correct Answer - C | |