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JEE • MATHEMATICS • PAIR OF STRAIGHT LINES

Pair of straight lines practice · Easy

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Showing 121-150 of 201 questions

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121 Easy
121 of 201
Application +4 | -1
If the equations, 12x2−10xy+2y2+11x−5y+k=012{x^2} - 10xy + 2{y^2} + 11x - 5y + k = 0 represents two straight lines, then the value of kk is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the value of k is found by applying the condition for a general second-degree equation to represent a pair of straight lines.
122 Easy
122 of 201
Application +4 | -1
The value of λ\lambda for which the equation 6x2+11xy−10y2+x+31y+λ=06 x^2+11 x y-10 y^2+x+31 y+\lambda=0 represents a pair of straight lines, is
View answer and explanation
Correct option: A
-15
Option A (-15) is correct because it satisfies the condition for the given equation to represent a pair of straight lines, where the determinant Δ=0\Delta = 0.
123 Easy
123 of 201
Application +4 | -1
If ‘θ\theta’ is the angle between the lines ax2+2hxy+by2=0,  a{x^2} + 2hxy + b{y^2} = 0,\; then angle between x2+2xysec  θ+y2=0{x^2} + 2xy\mathrm{sec\;\theta } + {y^2} = 0 is
View answer and explanation
Correct option: A
θ\theta
Option A (θ\theta) is correct because the angle between the second pair of lines simplifies to the same expression as the angle between the first pair.
124 Easy
124 of 201
Analyzing +4 | -1
Two of the straight lines given by 3x3+3x2y−3xy2+dy3=03 x^3+3 x^2 y-3 x y^2+d y^3=0 are at right angles, if dd equal to
View answer and explanation
Correct option: B
−3-3
Option B (-3) is correct because substituting the condition for perpendicular lines into the given equation yields d = -3.
125 Easy
125 of 201
Application +4 | -1
All chords of the curve 3x2−y2−2x+4y=03{x^2} - {y^2} - 2x + 4y = 0 which subtend a right angle at the origin, pass through the fix point
View answer and explanation
Correct option: B
(1,-2)
The correct answer is A (1,-2) because the equation of the chord is used to homogenize the given curve equation, and then the condition for the lines to be perpendicular is applied to find the fixed point.
126 Easy
126 of 201
Analyzing +4 | -1
The equation of line which is parallel to the line common to the pair of lines given by 3x2+xy−4y2=03 x^2+x y-4 y^2=0 and 6x2+11xy+4y2=06 x^2+11 x y+4 y^2=0 and at a distance of 2 units from it is
View answer and explanation
Correct option: C
3x+4y=103 x+4 y=10
Option C (3x+4y=103 x+4 y=10) is correct because it represents a line parallel to the common line of the given pair of lines and is at a distance of 2 units from it.
127 Easy
127 of 201
Application +4 | -1
The value of λ\lambda such that λ\lambda x2−10xy+12y2+5x−16y−3=0{x^2} - 10xy + 12{y^2} + 5x - 16y - 3 = 0 represent a pair of straight lines, is
View answer and explanation
Correct option: C
2
Option C (2) is correct because this value of λ\lambda makes the given equation represent a pair of straight lines.
128 Easy
128 of 201
Application +4 | -1
The pair of lines 3x2−4xy+3y2=0\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0 are rotated about the origin by π6\frac{\pi}{6} in the anti-clockwise sense.
The equation of the pair in the new position is
View answer and explanation
Correct option: C
3x2−xy=0\sqrt{3} x^2-x y=0
The correct answer is C (3x2−xy=0\sqrt{3} x^2-x y=0) because rotating the pair of lines by π6\frac{\pi}{6} results in the equation 3x2−xy=0\sqrt{3} x^2-x y=0.
129 Easy
129 of 201
Application +4 | -1
If the sum of the slopes of the lines given by x2+2cxy−y2=0x^2+2 c x y-y^2=0 is four times their product, then cc has the value
View answer and explanation
Correct option: A
−2-2
Option A (-2) is correct because it satisfies the condition where the sum of the slopes is four times their product for the given equation.
130 Easy
130 of 201
Understanding +4 | -1
The angle between the lines represented by x2−y2=0{x^2} - {y^2} = 0 is
View answer and explanation
Correct option: C
90∘90^\circ
Option C (90∘90^\circ) is correct because the equation x2−y2=0{x^2} - {y^2} = 0 represents two perpendicular lines.
131 Easy
131 of 201
Application +4 | -1
Two lines are given by (x−2y)2+λ(x−2y)=0(x-2 y)^2+\lambda(x-2 y)=0. The value of ∣λ∣|\lambda| so that the distance between them is 3 , is
View answer and explanation
Correct option: C
353 \sqrt{5}
The correct answer is C (353 \sqrt{5}) because the distance between the parallel lines is given as 3, and solving for ∣λ∣|\lambda| yields 353 \sqrt{5}.
132 Easy
132 of 201
Application +4 | -1
The equation of two straight lines through the point (x1,y1)\left(x_1, y_1\right) and perpendicular to the lines given by ax2+2hxy+by2=0,a x^2+2 h x y+b y^2=0, is
View answer and explanation
Correct option: A
b(x−x1)2−2h(x−x1)(y−y1)+a(y−y1)2=0b\left(x-x_1\right)^2-2 h\left(x-x_1\right)\left(y-y_1\right)+a\left(y-y_1\right)^2=0
Option A (b(x−x1)2−2h(x−x1)(y−y1)+a(y−y1)2=0b(x-x_1)^2-2 h(x-x_1)(y-y_1)+a(y-y_1)^2=0) is correct because it represents the equation of lines perpendicular to the given lines, shifted to pass through the point (x1, y1).
133 Easy
133 of 201
Application +4 | -1
The equation of pair of lines through origin and perpendicular to the pair of lines ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 is
View answer and explanation
Correct option: D
bx2−2hxy+ay2=0b x^2-2 h x y+a y^2=0
Option D (bx2−2hxy+ay2=0b x^2-2 h x y+a y^2=0) is correct because it represents the equation of a pair of lines perpendicular to the given pair of lines and passing through the origin.
134 Easy
134 of 201
Application +4 | -1
The area (in square unit) of the triangle formed by x+y+1=0x + y + 1 = 0 and the pair of straight lines x2−3xy+2y2=0{x^2} - 3xy + 2{y^2} = 0 is
View answer and explanation
Correct option: C
112\frac{1}{{12}}
Option C (112\frac{1}{{12}}) is correct because it represents the area of the triangle formed by the line and the pair of straight lines, calculated by finding the intersection points and using the area formula.
135 Easy
135 of 201
Application +4 | -1
The equation of the bisector of the angle between the straight lines 3x−4y+7=03 x-4 y+7=0 and 12x−5y−8=012 x-5 y-8=0, is
View answer and explanation
Correct option: D
21x+27y−131=021 x+27 y-131=0
Option D (21x+27y−131=021 x+27 y-131=0) is correct because it is the equation of one of the angle bisectors between the two given lines.

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136 Easy
136 of 201
Application +4 | -1
An equation of a straight line passing through the inter-section of the straight lines 3x−4y+1=03 x-4 y+1=0 and 5x+y−1=05 x+y-1=0 and making non-zero, equal intercepts on the axes is
View answer and explanation
Correct option: B
23x+23y=1123 x+23 y=11
Option D (4x+y=13,4x−y=74 x+y=13,4 x-y=7) is correct because these are the equations of the diagonals of the parallelogram formed by the given lines.
137 Easy
137 of 201
Application +4 | -1
The line x=ay+b,z=cy+dx=a y+b, z=c y+d and x=a′y+b′,z=c′y+d′x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime} are perpendicular if-
View answer and explanation
Correct option: D
aa′+cc′+1=0{aa}^{\prime}+{cc}^{\prime}+1=0
Option D ({aa}^{\'}+{cc}^{\'}+1=0{aa}^{\'}+{cc}^{\'}+1=0) is correct because the lines are perpendicular if the dot product of their direction ratios is zero, leading to the given condition.
138 Easy
138 of 201
Application +4 | -1
The angle between the straight lines x2−y2−2  x−1=0,{x^2} - {y^2} - 2\;x - 1 = 0, is
View answer and explanation
Correct option: A
90∘90^\circ
Option A (90∘90^\circ) is correct because the given equation represents a pair of perpendicular lines since the coefficient of x2x^2 plus the coefficient of y2y^2 equals zero.
139 Easy
139 of 201
Understanding +4 | -1
The joint equation of the straight lines x+y=1x + y = 1 and x−y=4,x - y = 4, is
View answer and explanation
Correct option: C
(x+y−1)(x−y−4)=0\left( {x + y - 1} \right)\left( {x - y - 4} \right) = 0
Option C ((x+y−1)(x−y−4)=0\left( {x + y - 1} \right)\left( {x - y - 4} \right) = 0) is correct because the joint equation is obtained by multiplying the individual equations after setting them equal to zero.
140 Easy
140 of 201
Application +4 | -1
The angle between straight lines passing through (1,0)(1,0) and parallel to straight lines 2x2−xy−6y2+7x+21y=152 x^2-x y-6 y^2+7 x+21 y=15 is
View answer and explanation
Correct option: C
tan⁡−1(7/4)\tan ^{-1}(7 / 4)
Option C (tan⁡−1(74)\tan^{{-1}}(\frac{{7}}{{4}})) is correct because it is the angle calculated using the formula for the angle between two lines represented by a second-degree equation.
141 Easy
141 of 201
Application +4 | -1
If the equation 2x2+7xy+3y2−9x−7y+k=02 x^2+7 x y+3 y^2-9 x-7 y+k=0 represents a pair of lines, then kk is equal to
View answer and explanation
Correct option: A
4
Option A (4) is correct because substituting the coefficients into the condition for a pair of straight lines and solving for kk yields k=4k=4.
142 Easy
142 of 201
Application +4 | -1
The lines joining the origin to the points of intersection of 3x2+λxy−4x+1=03 x^2+\lambda x y-4 x+1=0 and 2x+y−1=02 x+y-1=0 are at right angles for
View answer and explanation
Correct option: D
all values of λ\lambda
Option D (all values of λ\lambda) is correct because the homogeneous equation of the lines joining the origin to the intersection points results in a pair of perpendicular lines, independent of λ\lambda.
143 Easy
143 of 201
Application +4 | -1
If the equation kx2−2xy−y2−2x+2y=0k{x^2} - 2xy - {y^2} - 2x + 2y = 0 represents a pair of lines, then kk is equal to
View answer and explanation
Correct option: D
3
Option D (3) is correct because it satisfies the condition for the equation to represent a pair of lines.
144 Easy
144 of 201
Application +4 | -1
The line 2x−y=12x - y = 1 bisects angle between two lines. If equation of one line is y=x,y = x, then the equation of the other line is
View answer and explanation
Correct option: B
x−2y+1=0x - 2y + 1 = 0
Option B (2/3\sqrt {2/3}) is correct because it represents the length of each side of the equilateral triangle given the vertex and the base.
145 Easy
145 of 201
Application +4 | -1
The equation to the pair of straight lines through origin and perpendicular to lines ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 is
View answer and explanation
Correct option: C
bx2−2hxy+ay2=0b x^2-2 h x y+a y^2=0
Option C (bx2−2hxy+ay2=0b x^2-2 h x y+a y^2=0) is correct because it represents the equation of the pair of straight lines through the origin and perpendicular to the given lines.
146 Easy
146 of 201
Analyzing +4 | -1
The three lines whose combined equation is y3−4x2y=0y^3-4 x^2 y=0 form a triangle which is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the equation y3−4x2y=0y^3-4 x^2 y=0 represents three concurrent lines passing through the origin, and thus does not form a triangle.
147 Easy
147 of 201
Application +4 | -1
The point of intersection of lines represented by 3x2+10xy+3 x^2+10 x y+ 8y2+14x+22y+15=08 y^2+14 x+22 y+15=0 is
View answer and explanation
Correct option: B
(1,−2)(1,-2)
Option B ((1,-2)) is correct because it is the point of intersection found by solving the partial derivative equations of the given equation with respect to x and y.
148 Easy
148 of 201
Application +4 | -1
Distance between the pair of lines represented by the equation x2−6xy+9y2+3x−9y−4=0,{x^2} - 6xy + 9{y^2} + 3x - 9y - 4 = 0, is
View answer and explanation
Correct option: C
52\sqrt {\frac{5}{2}}
Option C (52\sqrt {\frac{{5}}{{2}}}) is correct because the distance between the parallel lines is calculated using the formula 2g2−aca(a+b)2\sqrt {\frac{{{{g^2} - ac}}}{{a\left( {a + b} \right)}}}.
149 Easy
149 of 201
Application +4 | -1
The equation of the straight lines through the point (x1,y1)\left(x_1, y_1\right) and parallel to the lines given by ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0, is
View answer and explanation
Correct option: C
b(y−y1)2+2h(x−x1)(y−y1)+a(x−x1)2=0b\left(y-y_1\right)^2+2 h\left(x-x_1\right)\left(y-y_1\right)+a\left(x-x_1\right)^2=0
The correct answer is D (None of these) because the equation of the lines parallel to the given lines and passing through (x1, y1) is a(x−x1)2+2h(x−x1)(y−y1)+b(y−y1)2=0a(x-x_1)^2+2h(x-x_1)(y-y_1)+b(y-y_1)^2=0. The provided answer in the solution steps is also incorrect.
150 Easy
150 of 201
Application +4 | -1
The equation of the pair of straight lines perpendicular of the pair 2x2+3xy+2y2+10x+5y=02{x^2} + 3xy + 2{y^2} + 10x + 5y = 0 and passing through the origin, is
View answer and explanation
Correct option: B
2x2−3xy+2y2=02{x^2} - 3xy + 2{y^2} = 0
Option B (2x2−3xy+2y2=02{x^2} - 3xy + 2{y^2} = 0) is correct because it represents the equation of the pair of straight lines perpendicular to the given pair and passing through the origin.
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