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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Mixed difficulty

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61 Medium
61 of 555
Application +4 | -1
The condition that the pair of straight line joining the origin to the intersections of the line y=mx+cy=m x+c and the circle x2+y2=a2x^2+y^2=a^2 is at the right angles, is 
View answer and explanation
Correct option: B
2c2=a2(1+m2)2 c^2=a^2\left(1+m^2\right)
Option B (2c2=a2(1+m2)2 c^2=a^2(1+m^2)) is correct because it represents the condition for the pair of straight lines joining the origin to the intersection of the given line and circle to be at right angles.
62 Easy
62 of 555
Analyzing +4 | -1
Area of quadrilateral whose vertices are (2, 3), (3, 4), (4, 5) and(5, 6), is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the area of the quadrilateral with the given vertices is zero, indicating that the points are collinear.
63 Medium
63 of 555
Application +4 | -1
y-intercept of line passes through (2, 2) and is perpendicular to the line 3x+y=33x + y = 3, is
View answer and explanation
Correct option: D
43\frac{4}{3}
The correct answer is D (4/3) because it's the y-intercept of the line that passes through (2,2) and is perpendicular to the given line.
64 Easy
64 of 555
Application +4 | -1
A square of area 25 sq unit is formed by taking two sides as 3x+4y=k13x + 4y = {k_1} and 3x+4y=k23x + 4y = {k_2}, then ∣k1−k2∣\left| {{k_1} - {k_2}} \right| is
View answer and explanation
Correct option: C
25
Option D (25) is correct because the distance between the parallel lines, which forms the side of the square, is equal to the side length of the square, and the absolute difference between the constants is 25.
65 Easy
65 of 555
Application +4 | -1
If A(−5,  0)A\left( { - 5,\;0} \right) and B(3,  0)B\left( {3,\;0} \right) are two vertices of a triangle ABCABC. Its area is 20 sq cm. The vertex CC lies on the line x−y=2x - y = 2. The coordinates of CC are
View answer and explanation
Correct option: D
(−3,  −5)or  (7,  5)\left( { - 3,\; - 5} \right)\mathrm{or}\;\left( {7,\;5} \right)
Option D ((-3, -5) or (7, 5)) is correct because these coordinates satisfy both the area condition and the line equation.
66 Medium
66 of 555
Application +4 | -1
The equation of the obtuse angle bisector of lines 12x−5y+7=012 x-5 y+7=0 and 3y−4x−1=03 y-4 x-1=0 is
View answer and explanation
Correct option: A
4x+7y+11=04 x+7 y+11=0
Option A (4x+7y+11=04 x+7 y+11=0) is correct because it is the equation of the obtuse angle bisector, determined by checking the sign of a1a2+b1b2a_1a_2 + b_1b_2.
67 Medium
67 of 555
Analyzing +4 | -1
A variable straight line drawn through the point of intersection of lines xa+yb=1\frac{x}{a}+\frac{y}{b}=1 and xb+ya=1\frac{x}{b}+\frac{y}{a}=1, meets the coordinate axes in AA and BB. Then, the locus of the mid-point of ABA B is
View answer and explanation
Correct option: D
2xy(a+b)=ab(x+y)2 x y(a+b)=a b(x+y)
Option D (2xy(a+b)=ab(x+y)2 x y(a+b)=a b(x+y)) is correct because it represents the locus of the midpoint of AB, derived from the intersection of the given lines and the variable line.
68 Easy
68 of 555
Application +4 | -1
The distance of the line 2x+3y−5=02 x+3 y-5=0 from the point (3,5)(3,5) along the line 5x−3y=05 x-3 y=0 in units is
View answer and explanation
Correct option: C
43421\frac{4 \sqrt{34}}{21}
Option C (43421\frac{4 \sqrt{34}}{21}) is correct because it represents the distance of the line from the point along the given line.
69 Easy
69 of 555
Application +4 | -1
If the pair of straight lines xy−x−y+1=0xy - x - y + 1 = 0 and the line ax+2y−3=0ax + 2y - 3 = 0 are concurrent, then aa is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because the concurrency condition requires the intersection point of the first two lines to lie on the third line, leading to a=1.
70 Medium
70 of 555
Analyzing +4 | -1
If the straight lines a1x+b1y+c1=0a_1 x+b_1 y+c_1=0, a1x+b1y+c2=0,a2x+b2y+d1=0a_1 x+b_1 y+c_2=0, a_2 x+b_2 y+d_1=0 and a2x+b2y+d2=0a_2 x+b_2 y+d_2=0 are the sides of rhombus, then
View answer and explanation
Correct option: A
(a22+b22)(c1−c2)2=(a12+b12)(d1−d2)2\left(a_2^2+b_2^2\right)\left(c_1-c_2\right)^2=\left(a_1^2+b_1^2\right)\left(d_1-d_2\right)^2
Option A is correct because it equates the ratio of the distance between parallel lines to be equal, which is a property of a rhombus.
71 Easy
71 of 555
Analyzing +4 | -1
If the lines represented by the equation ax2+2hxy+by2+2gx+2fy+c=0a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 are equidistant from the origin, then
View answer and explanation
Correct option: A
f4−g4=c(bf2−ag2){f^4} - {g^4} = c\left( {b{f^2} - a{g^2}} \right)
The correct answer is f4−g4=c(bf2−ag2){f^4} - {g^4} = c\left( {b{f^2} - a{g^2}} \right) because the condition for the lines to be equidistant from the origin leads to this relationship between the coefficients.
72 Medium
72 of 555
Application +4 | -1
Let P=(-1,0),Q=(0,0) and R=(3,  33)R = \left( {3,\;3\sqrt 3 } \right) be three points. The equation of the bisector of the angle PQR is
View answer and explanation
Correct option: A
3x+y=0\sqrt 3 x + y = 0
Option A (3x+y=0\sqrt 3 x + y = 0) is correct because the bisector's equation is derived from the angle between the lines and the slope.
73 Easy
73 of 555
Understanding +4 | -1
Condition that the two lines represented by the equation ax2+2hxy+by2=0a x^2+2 h x y+b y^2=0 to be perpendicular is
View answer and explanation
Correct option: B
a=−ba=-b
Option B (a=−ba=-b) is correct because for perpendicular lines, the sum of the coefficients of x² and y² must be zero.
74 Easy
74 of 555
Application +4 | -1
The equation of a line which is equidistant from the lines x=−4x=-4 and x=8,x=8, is
View answer and explanation
Correct option: A
x=2x=2
Option A (x=2) is correct because the line equidistant from x=-4 and x=8 is a vertical line located at the average of the two x-values, which is x=2.
75 Medium
75 of 555
Application +4 | -1
Area of the triangle formed by the lines y=2x,y=3xy = 2x,y = 3x and y=5y = 5 is equal to (in square unit)
View answer and explanation
Correct option: B
2512\frac{{25}}{{12}}
Option B 2512\frac{{25}}{{12}} is correct because it is the area of the triangle calculated using the coordinates of the intersection points.

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76 Easy
76 of 555
Application +4 | -1
If the lines x=a+m,  y=−2  x = a + m,\;y = - 2\; and y=mxy = mx are concurrent, then least value of ∣a∣\left| a \right| is
View answer and explanation
Correct option: C
222\sqrt 2
Option C (222\sqrt 2) is correct because it is the least value of |a| that ensures the existence of real solutions for m.
77 Easy
77 of 555
Understanding +4 | -1
If ∣x1y11x2y21x3y31∣=0\left| {\begin{array}{cccccccccccccccccccc} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0, then the points (x1,y1),  (x2,  y2)\left( {{x_1},{y_1}} \right),\;\left( {{x_2},\;{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right) are
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because if the determinant is zero, the points are collinear, not vertices of any triangle.
78 Easy
78 of 555
Application +4 | -1
The image of the point (−1,  3)\left( { - 1,\;3} \right) by the line x−y=0x - y = 0 is
View answer and explanation
Correct option: A
(3,−1)\left( {3, - 1} \right)
Option A ((3, -1)) is correct because the image of a point (x, y) across the line x - y = 0 (or y = x) is (y, x).
79 Easy
79 of 555
Analyzing +4 | -1
Straight lines xa+yb=1,xb+ya=1,xa+yb=2\frac{x}{a}+\frac{y}{b}=1, \frac{x}{b}+\frac{y}{a}=1, \frac{x}{a}+\frac{y}{b}=2 and xb+ya=2\frac{x}{b}+\frac{y}{a}=2 form a rhombus of area (in square units)
View answer and explanation
Correct option: D
a2b2∣a2−b2∣\frac{a^2 b^2}{\left|a^2-b^2\right|}
Option D (a2b2∣a2−b2∣\frac{{a^2 b^2}}{{\left|a^2-b^2\right|}}) is correct because it represents the area of the rhombus formed by the given lines, calculated using the determinant method.
80 Easy
80 of 555
Application +4 | -1
The number of integral values of m, for which the x-coordinate of the point of intersection of the lines 3x+4y=93x + 4y = 9 and y=mx+1y = mx + 1 is also an integer, is
View answer and explanation
Correct option: A
2
The correct answer is A (2) because there are two integer values of m that make the x-coordinate of the intersection point an integer.
81 Easy
81 of 555
Application +4 | -1
The bisector of the acute angle formed between the lines 4x−3y+7=04x - 3y + 7 = 0 and 3x−4y+14=03x - 4y + 14 = 0 has the equation
View answer and explanation
Correct option: C
x−y+3=0x - y + 3 = 0
Option C (x - y + 3 = 0) is correct because it represents the bisector of the acute angle between the two lines, determined by checking the sign.
82 Medium
82 of 555
Application +4 | -1
If 3a−2b+5c=03 a-2 b+5 c=0, family of the lines ax+by+c=0a x+b y+c=0 are always concurrent at a point whose coordinates are
View answer and explanation
Correct option: B
(35,−25)\left(\frac{3}{5},-\frac{2}{5}\right)
Option B ((35,−25)\left(\frac{3}{5},-\frac{2}{5}\right)) is correct because the family of lines passes through the point that satisfies the given condition 3a−2b+5c=03a - 2b + 5c = 0.
83 Medium
83 of 555
Application +4 | -1
The lines x+2y+3=0,x+2y−7=0x+2 y+3=0, x+2 y-7=0 and 2x−y−4=02 x-y-4=0 are the sides of a square. Equation of the remaining side of the square can be
View answer and explanation
Correct option: A
2x−y−14=02 x-y-14=0
Option A (2x−y−14=02x-y-14=0) is correct because it represents a line parallel to the given parallel side of the square and at the correct distance to form a square.
84 Easy
84 of 555
Application +4 | -1
If a line joining two points A(2,0)A(2,0) and B(3,1)B(3,1) is rotaled about AA in anticlockwise direction through an angle 15∘15^{\circ} such that the point BB goes to CC in the new position, then the coordinates of CC are
View answer and explanation
Correct option: D
(2−12,−32)\left(2-\frac{1}{\sqrt{2}},-\frac{\sqrt{3}}{2}\right)
Option D (2−12,−32)\left(2-\frac{1}{\sqrt{2}},-\frac{\sqrt{3}}{2}\right) is correct because it represents the coordinates of the point C after rotating point B about A by 15 degrees in the anticlockwise direction.
85 Easy
85 of 555
Application +4 | -1
The equation of line bisecting perpendicularly the segment joining the points (-4,6) and (8, 8), is
View answer and explanation
Correct option: B
6x+y−19=06x + y - 19 = 0
Option A (20 sq. units) is correct because it is the area of the rectangle calculated using the distance between the given vertices and the perpendicular distance from a vertex to the given side.
86 Easy
86 of 555
Application +4 | -1
Find the image of the point P(4,−13)P(4,-13) in the line mirror 5x+y+6=05 x+y+6=0, is
View answer and explanation
Correct option: C
(−1,−14)(-1,-14)
Option C is correct. The image of the point P(4,−13)P(4, -13) in the line 5x+y+6=05x + y + 6 = 0 is found using the formula for the image of a point in a line, resulting in (−1,−14)(-1, -14).
87 Easy
87 of 555
Application +4 | -1
If the slope of one line is double the other in line pair x2+6xy+by2=0x^2+6 x y+b y^2=0, then b=b=?
View answer and explanation
Correct option: D
8
Option D (8) is correct because it satisfies the condition that one slope is double the other in the given pair of lines.
88 Hard
88 of 555
Application +4 | -1
The lines xcos⁡α+ysin⁡α=p1x\cos \alpha + y\sin \alpha = {p_1} and xcos⁡β+ysin⁡β=p2x\cos \beta + y\sin \beta = {p_2} will be perpendicular, if
View answer and explanation
Correct option: C
∣α−β∣=π2\left| {\alpha - \beta } \right| = \frac{\pi }{2}
Option C (∣α−β∣=π2\left| {\alpha - \beta } \right| = \frac{\pi }{2}) is correct because the lines are perpendicular when the difference in their angles is π2\frac{\pi }{2}.
89 Easy
89 of 555
Application +4 | -1
If the sum of the distance of a point PP from two perpendicular lines in a plane is 1, then the locus of PP is a
View answer and explanation
Correct option: A
Rhombus
Option A (Rhombus) is correct because the locus of a point with a constant sum of distances from two perpendicular lines forms a rhombus.
90 Easy
90 of 555
Understanding +4 | -1
Angles made by the lines represented by the equation xy+y=0x y+y=0 with yy-axis are
View answer and explanation
Correct option: A
0∘0^{\circ} and 90∘90^{\circ}
Option A (0∘0^{\circ} and 90∘90^{\circ}) is correct because the lines represented are y=0 and x=-1, making angles of 90 and 0 degrees with the y-axis respectively.
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