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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Easy

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Showing 361-390 of 408 questions

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361 Easy
361 of 408
Analyzing +4 | -1
The lines 2  x+y−1=0,  ax+3  y−3=02\;x + y - 1 = 0,\;ax + 3\;y - 3 = 0 and 3x+2  y−2=03x + 2\;y - 2 = 0 are concurrent for
View answer and explanation
Correct option: A
All aa
Option A (All aa) is correct because the determinant of the coefficients is always zero, indicating concurrency regardless of the value of 'a'.
362 Easy
362 of 408
Application +4 | -1
The least value of ∣k∣|k|, so that the lines x=k+m,y=−2x=k+m, y=-2 and y=mxy=m x are concurrent, is
View answer and explanation
Correct option: A
222 \sqrt{2}
Option A (222 \sqrt{{2}}) is correct because it represents the least value of |k| for which the lines are concurrent, derived from the discriminant condition for real roots.
363 Easy
363 of 408
Application +4 | -1
If the sum of distances from a point P on two mutually perpendicular straight lines is 1 unit, then the locus of P is
View answer and explanation
Correct option: D
A straight line
Option D (A straight line) is correct because the locus of P, defined by |x| + |y| = 1, represents a square, which consists of straight line segments.
364 Easy
364 of 408
Application +4 | -1
If the points (a,a2)\left(a, a^2\right) and (1,2)(1,2) lie in the same angular region between the lines 3x+4y−1=03 x+4 y-1=0 and 2x+y−3=02 x+y-3=0, then
View answer and explanation
Correct option: A
a<−3a<-3 or, a>1a>1
Option A (a<−3a<-3 or, a>1a>1) is correct because these values of 'a' ensure that the points (a, a²) and (1,2) lie in the same angular region between the two lines.
365 Easy
365 of 408
Understanding +4 | -1
If the distance of any point (x,y)(x, y) from the origin is defined as d(x,y)=max⁡{∣x∣,∣y∣}d(x, y)=\max \{|x|,|y|\}, then the locus of the point (x,y)(x, y), where d(x,y)=1d(x, y)=1, is
View answer and explanation
Correct option: B
a square
Option B (a square) is correct because the given distance function defines a square centered at the origin with side length 2.
366 Easy
366 of 408
Application +4 | -1
The locus of the point of intersection of the limes xcos⁡α+ysin⁡α=ax \cos \alpha+y \sin \alpha=a and xsin⁡α−ycos⁡α=bx \sin \alpha-y \cos \alpha=b, where α\alpha is a variable, is
View answer and explanation
Correct option: C
x2+y2=a2+b2x^2+y^2=a^2+b^2
Option C is correct. The locus of the point of intersection of the lines xcos⁡α+ysin⁡α=ax \cos \alpha+y \sin \alpha=a and xsin⁡α−ycos⁡α=bx \sin \alpha-y \cos \alpha=b, where α\alpha is a variable, is x2+y2=a2+b2x^2+y^2=a^2+b^2.
367 Easy
367 of 408
Application +4 | -1
Point where all lines of family x(a+2b)+y(a+3b)=a+bx\left( {a + 2b} \right) + y\left( {a + 3b} \right) = a + b are concurrent is
View answer and explanation
Correct option: B
(2,−1)(2,-1)
Option B (2,-1) is correct because it satisfies the equation for all values of a and b, making it the point of concurrency.
368 Easy
368 of 408
Application +4 | -1
The point on the axis of xx, whose perpendicular distance from the straight line xa+yb=1\frac{x}{a} + \frac{y}{b} = 1 is aa, are
View answer and explanation
Correct option: B
(ab(  b±a2+b2),  0)\left( {\frac{a}{b}\left( {\;b \pm \sqrt {{a^2} + {b^2}} } \right),\;0} \right)
The correct answer is B ((ab\frac{a}{b}(b ± √(a²+b²)), 0)) because the perpendicular distance from a point on the x-axis to the given line is calculated and set equal to 'a'.
369 Easy
369 of 408
Application +4 | -1
The points (1, 3) and (5, 1) are the opposite vertices of a rectangle. The other two vertices lie on the line y=2x+cy = 2x + c, then the value of c will be
View answer and explanation
Correct option: B
−4- 4
Option B (-4) is correct because the midpoint of the diagonal lies on the given line, and substituting the midpoint coordinates into the line equation gives c = -4.
370 Easy
370 of 408
Understanding +4 | -1
If u=a1x+b1y+c1=0,v=a2x+b2y+c2=0u=a_1 x+b_1 y+c_1=0, v=a_2 x+b_2 y+c_2=0 and a1a2=b1b2=c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}, then u+kv=0u+k v=0 represents
View answer and explanation
Correct option: A
u=0u=0
Option A (u=0u=0) is correct because if a1a2=b1b2=c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}, then the two lines are identical, and u+kv=0u+kv=0 represents the same line.
371 Easy
371 of 408
Application +4 | -1
Points on the line x+y=4 which are equidistant from the lines ∣x∣=∣y∣,\left| x \right| = \left| y \right|, are
View answer and explanation
Correct option: A
(4,  0),(0,  4)\left( {4,\;0} \right),\left( {0,\;4} \right)
Option A ((4,  0),(0,  4)\left( {4,\;0} \right),\left( {0,\;4} \right)) is correct because these points on the line x+y=4 are equidistant from the lines |x|=|y|.
372 Easy
372 of 408
Application +4 | -1
A ray of light coming from the point (1,2)(1,2) is reflected at a point AA on the xx-axis and then passes through the point (5,3)(5,3). The coordinates of the point AA are (x,y)(x, y). Find 5x+y5 x+y.
View answer and explanation
Correct option: A
13
The correct answer is A (13) because it is the value of 5x+y after finding the coordinates of point A using the reflection property.
373 Easy
373 of 408
Application +4 | -1
The equation of the straight line through the intersection of line 2x+y=12 x+y=1 and 3x+2y=53 x+2 y=5 and passing through the origin, is
View answer and explanation
Correct option: A
7x+3y=07 x+3 y=0
Option A (7x+3y=07 x+3 y=0) is correct because it represents the equation of a line passing through the intersection of the given lines and the origin.
374 Easy
374 of 408
Application +4 | -1
If A(3,  5),  B(−5,  −4),  C(7,  10)A\left( {3,\;5} \right),\;B\left( { - 5,\; - 4} \right),\;C\left( {7,\;10} \right) are the vertices of a parallelogram, taken in the order, then the coordinates of the fourth vertex are
View answer and explanation
Correct option: B
(15, 19)
Option A (15, 19) is correct because in a parallelogram, the midpoints of the diagonals coincide. Therefore, the midpoint of AC must be the same as the midpoint of BD, where D is the fourth vertex.
375 Easy
375 of 408
Application +4 | -1
If xcos⁡α+ysin⁡α=−sin⁡αtan⁡αx \cos \alpha+y \sin \alpha=-\sin \alpha \tan \alpha be the equation of a line, then the length of the perpendiculars on the line from the points (a2,2a),(ab,a+b)\left(a^2, 2 a\right),(a b, a+b) and (b2,2b)\left(b^2, 2 b\right) are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because the lengths of the perpendiculars form a geometric progression, as shown by the relationship l1l3=l22l_1 l_3 = l_2^2.

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376 Easy
376 of 408
Application +4 | -1
Equation of line passing through point of intersection of x+y+2=0x+y+2=0 and x−y+4=0x-y+4=0 and having xx-intercept =0=0
View answer and explanation
Correct option: D
x+3y=0x+3 y=0
Option D (x+3y=0x+3y=0) is correct because it passes through the intersection of the given lines and has an x-intercept of 0, meaning it passes through the origin.
377 Easy
377 of 408
Application +4 | -1
ABCABC is an isosceles triangle if the coordinates of the base are B(1,  3)B\left( {1,\;3} \right) and C(−2,  7)C\left( { - 2,\;7} \right), the coordinates of vertex AA can be
View answer and explanation
Correct option: C
(56,  6)\left( {\frac{5}{6},\;6} \right)
Option C (56,  6)\left( {\frac{5}{6},\;6} \right) is correct because it satisfies the condition that vertex A is equidistant from B and C in an isosceles triangle.
378 Easy
378 of 408
Analyzing +4 | -1
The circumcentre of the triangle formed by the lines y=x,  y=2xy = x,\;y = 2x and y=3x+4y = 3x + 4 is
View answer and explanation
Correct option: B
(6,  −8)\left( {6,\; - 8} \right)
Option B ((6, -8)) is correct because it represents the circumcenter of the triangle formed by the given lines, found by solving equations derived from the circumcenter's properties.
379 Easy
379 of 408
Application +4 | -1
The number of lines that are parallel to 2  x+6  y+7=02\;x + 6\;y + 7 = 0 and have an intercept of length 10 between the coordinate axes, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because there are two lines that satisfy the given conditions of being parallel and having a specific intercept length.
380 Easy
380 of 408
Application +4 | -1
If the straight line ax+by+c=0ax + by + c = 0 always passes through (1,-2) then a,b,c are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because substituting the point (1, -2) into the line equation results in a relationship that shows a, b, and c are in arithmetic progression.
381 Easy
381 of 408
Application +4 | -1
The image of the origin with reference to the line 4x+3y−25=0,  4x + 3y - 25 = 0,\; is
View answer and explanation
Correct option: B
(8,6)
The correct answer is B (8,6) because the image of the origin with respect to the line 4x + 3y - 25 = 0 is found using the reflection formula.
382 Easy
382 of 408
Application +4 | -1
The point of concurrence of the lines ax+by+c=0ax + by + c = 0 and a,b,c satisfy the relation 3a+2b+4c=03a + 2b + 4c = 0 is
View answer and explanation
Correct option: C
(34,12)\left( {\frac{3}{4},\frac{1}{2}} \right)
Option D ((34,12)\left( {\frac{3}{4},\frac{1}{2}} \right)) is correct because it satisfies the given relation 3a+2b+4c=03a + 2b + 4c = 0.
383 Easy
383 of 408
Application +4 | -1
If a circle is inscribed in an equilateral triangle of side aa, then area of the square inscribed in the circle is
View answer and explanation
Correct option: A
a26\frac{{{a^2}}}{6}
Option A (a26\frac{{{a^2}}}{6}) is correct because it is derived from calculating the inradius of the equilateral triangle and then finding the side of the inscribed square.
384 Easy
384 of 408
Application +4 | -1
If the lines 3x−4y−13=0,8x−11y−33=03 x-4 y-13=0,8 x-11 y-33=0 and 2x−3y+λ=02 x-3 y+\lambda=0 are concurrent, then λ=\lambda=
View answer and explanation
Correct option: C
−7-7
Option C (-7) is correct because the determinant of the coefficients of the concurrent lines must be zero, leading to λ = -7.
385 Easy
385 of 408
Application +4 | -1
Family of lines xsec⁡2θ+ytan⁡2θ−2=0x{\sec ^2}\theta + y{\tan ^2}\theta - 2 = 0 for different θ\theta real , is
View answer and explanation
Correct option: C
Concurrent at (2,−2)\left( {2, - 2} \right)
Option C ((2,−2)\left( {2, - 2} \right)) is correct because the family of lines passes through the intersection of two lines derived from the given equation.
386 Easy
386 of 408
Application +4 | -1
If points (a2,0),(0,b2)\left(a^2, 0\right),\left(0, b^2\right) and (1,1)(1,1) are collinear, then
View answer and explanation
Correct option: A
1a2+1b2=1\frac{1}{a^2}+\frac{1}{b^2}=1
Option A (1a2+1b2=1\frac{1}{a^2}+ \frac{1}{b^2}=1) is correct because it represents the condition for the three points to be collinear.
387 Easy
387 of 408
Application +4 | -1
The length of perpendicular from the point (acos⁡α,  asin⁡α)\left( {a\cos \mathrm{\alpha },\;a\sin \alpha } \right) upon the straight line y=xtan⁡α+c,c>0y = x\tan \alpha + c,c > 0 is
View answer and explanation
Correct option: C
ccos⁡αc\cos \mathrm{\alpha }
Option C (ccos⁡αc\cos \mathrm{\alpha }) is correct because it is the perpendicular distance calculated from the point to the line using the perpendicular distance formula.
388 Easy
388 of 408
Application +4 | -1
If the algebraic sum of the perpendicular distances from the points (2,0),(0,2)(2,0),(0,2) and (1,1)(1,1) to a variable straight line be zero, then the line passes through the point
View answer and explanation
Correct option: B
(1,1)(1,1)
Option B ((1,1)) is correct because substituting (1,1) into the derived equation satisfies the condition that the algebraic sum of perpendicular distances is zero.
389 Easy
389 of 408
Application +4 | -1
If the points (1,2) and (3,4) were to be on the same side of the line 3x−5y+a=0,3x - 5y + a = 0, then
View answer and explanation
Correct option: D
a<7a < 7 or a>11a > 11
Option D (a<7a < 7 or a>11a > 11) is correct because the product of the expressions obtained by substituting the points into the line's equation must be positive for the points to be on the same side.
390 Easy
390 of 408
Analyzing +4 | -1
The coordinates of the point PP on the line 2x+3y+1=02 x+3 y+1=0, such that ∣PA−PB∣|P A-P B| is maximum, where AA is (2,0)(2,0) and BB is (0,2)(0,2) is
View answer and explanation
Correct option: B
(7,−5)(7,-5)
Option B ((7,-5)) is correct because the maximum value of |PA - PB| occurs when P, A, and B are collinear, and solving the equations of the lines gives the coordinates of P.
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