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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Mixed difficulty

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391 Easy
391 of 555
Application +4 | -1
The straight line through the point of intersection of ax+by+c=0a x+b y+c=0 and a′x+b′y+c′=0a^{\prime} x+b^{\prime} y+c^{\prime}=0 are parallel to the yy-axis has the equation
View answer and explanation
Correct option: A
x(ab′−a′b)+(cb′−c′b)=0x\left(a b^{\prime}-a^{\prime} b\right)+\left(c b^{\prime}-c^{\prime} b\right)=0
Option A is correct. The equation of the straight line through the intersection of ax+by+c=0ax+by+c=0 and a′x+b′y+c′=0a'x+b'y+c'=0 and parallel to the y-axis is x(ab′−a′b)+(cb′−c′b)=0x(ab'-a'b)+(cb'-c'b)=0.
392 Medium
392 of 555
Application +4 | -1
The angle between the lines represented by the equation 4x2−24xy+11y2=04 x^2-24 x y+11 y^2=0 are
View answer and explanation
Correct option: C
tan⁡−143,tan⁡−1(−43)\tan ^{-1} \frac{4}{3}, \tan ^{-1}\left(-\frac{4}{3}\right)
Option C (tan⁡−143,tan⁡−1(−43)\tan ^{-1} \frac{4}{3}, \tan ^{-1}\left(-\frac{4}{3}\right)) is correct because it is calculated using the formula for the angle between lines represented by a homogeneous equation of degree 2.
393 Medium
393 of 555
Application +4 | -1
The equation of the line through (2,3)(2,3), so that the segment of the line intercepted between the axes is bisected at this point, is
View answer and explanation
Correct option: B
3x+2y=123 x+2 y=12
Option B (3x+2y=123x+2y=12) is correct because the line passes through (2,3) and the segment intercepted between the axes is bisected at this point.
394 Easy
394 of 555
Application +4 | -1
Point on line x+y=4x+y=4 which is at a unit distance from the line 4x+3y−11=04 x+3 y-11=0 is
View answer and explanation
Correct option: A
(4,0)(4,0)
Option A, (4,0), is one of the correct points on the line x+y=4 that is at a unit distance from the line 4x+3y-11=0.
395 Easy
395 of 555
Application +4 | -1
The coordinates of the foot of the perpendicular from the point (2,4) on the line x+y=1x + y = 1 are
View answer and explanation
Correct option: B
(−1/2,  3/2)\left( { - 1/2,\;3/2} \right)
Option B ((-1/2, 3/2)) is correct because it represents the coordinates of the foot of the perpendicular from the point (2,4) to the line x + y = 1.
396 Easy
396 of 555
Analyzing +4 | -1
Let AA be the fixed point (0,4)(0,4) and B(2t,0)B(2 t, 0) be a moving point. Let MM be the mid-point of ABA B and let the perpendicular bisector of ABA B meet the yy-axis at RR. The locus of the mid-point PP of MRM R, is
View answer and explanation
Correct option: C
y+x2=2y+x^2=2
Option C (y+x2=2y+x^2=2) is correct because it represents the locus of the midpoint P, derived from the given geometric conditions.
397 Easy
397 of 555
Application +4 | -1
The image of the origin with reference to the line 4x+3y−25=04x + 3y - 25 = 0 is
View answer and explanation
Correct option: B
(8,6)
Option B (8,6) is correct because it is found by using the reflection formula with respect to the given line.
398 Easy
398 of 555
Application +4 | -1
The equation of the line bisecting perpendicularly the segment joining the points (-4,6) and (8,8) is
View answer and explanation
Correct option: A
6x+y−19=06x + y - 19 = 0
Option B (6x+y−19=06x + y - 19 = 0) is correct because it represents the equation of the line that bisects perpendicularly the segment joining the given points.
399 Easy
399 of 555
Application +4 | -1
The distance between the lines 4  x+3  y=114\;x + 3\;y = 11 and 8  x+6  y=15,8\;x + 6\;y = 15, is
View answer and explanation
Correct option: C
07-Oct
Option C (0.7) is correct because it is the perpendicular distance between the two parallel lines after making the coefficients of x and y equal.
400 Easy
400 of 555
Application +4 | -1
The intercept made by a line on YY-axis is double to the intercept made by it on XX-axis, if it passes through (1,2)(1,2), then its equation is
View answer and explanation
Correct option: A
2x+y=42 x+y=4
Option A (2x+y=42x+y=4) is correct because it satisfies the given conditions: the y-intercept is double the x-intercept, and the line passes through the point (1, 2).
401 Medium
401 of 555
Application +4 | -1
The slopes of the lines joining the origin to the points of trisection of the portion of the line 3x+y=123x + y = 12 intercepted between the axis are the roots of the equation
View answer and explanation
Correct option: A
2x2−15x+18=02{x^2} - 15x + 18 = 0
Option A (2x2−15x+18=02{x^2} - 15x + 18 = 0) is correct because it represents the quadratic equation whose roots are the slopes of the lines joining the origin to the points of trisection of the given line segment.
402 Easy
402 of 555
Analyzing +4 | -1
The lines x+y=∣a∣x+y=|a| and ax−y=1a x-y=1 intersect each other in the first quadrant. Then the set of all possible values of a is the interval:
View answer and explanation
Correct option: A
[1,∞)[1, \infty)
Option A ([1,∞)[1, \infty)) is correct because the intersection point lies in the first quadrant only if a≥1a \ge 1.
403 Easy
403 of 555
Application +4 | -1
A line forms a triangle of area 54354 \sqrt{3} sq-units with the coordinate axes. If the perpendicular drawn from the origin to the line makes an angle of 60∘60^{\circ} with the xx-axis, then the equation of the line is
View answer and explanation
Correct option: C
x+3y=18x+\sqrt{3} y=18
Option C (x+3y=18x+\sqrt{3} y=18) is correct because it satisfies the given conditions of area and angle of the perpendicular from the origin.
404 Easy
404 of 555
Application +4 | -1
If p and qp\,\mathrm{and}\,q are the lengths of the perpendiculars from the origin on the lines, x cos⁡ec α − y sec⁡ α = k cot⁡ 2α and x sin⁡ α + y cos⁡ α = k sin⁡ 2α,x\,\cos ec\,\alpha \, - \,y\,\sec \,\alpha \, = \,k\,\cot \,2\alpha \,\mathrm{and}\,x\,\sin \,\alpha \, + \,y\,\cos \,\alpha \, = \,k\,\sin \,2\alpha , respectively then k2{k^2} is equal to
View answer and explanation
Correct option: A
4p2+q24{p^2} + {q^2}
Option D (p2+4q2{p^2} + 4{q^2}) is correct because by expressing p and q in terms of k and trigonometric functions, and then using the identity sin⁡2+cos⁡2=1{\sin^2} + {\cos^2} = 1, we can derive the relationship k2=4p2+q2{k^2} = 4{p^2} + {q^2}.
405 Easy
405 of 555
Application +4 | -1
The equation of straight line equally inclined to the axes and equidistance from the points (1,  −2)\left( {1,\; - 2} \right) and (3,4) is ax+by+c=0ax + by + c = 0, where
View answer and explanation
Correct option: C
a=1,  b=1,  c=−3a = 1,\;b = 1,\;c = - 3
Option B (3, -4) is correct because it represents the point on the line closest to the origin, which is the foot of the perpendicular from the origin to the line.

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406 Easy
406 of 555
Application +4 | -1
If the lines 3x+4y+1=0,  5x+λy+3=03x + 4y + 1 = 0,\;5x + \lambda y + 3 = 0 and 2x+y−1=02x + y - 1 = 0 are concurrent, then λ\lambda is equal to
View answer and explanation
Correct option: B
8
Option B (8) is correct because it satisfies the condition for concurrency of the three given lines, found by solving for the intersection point of two lines and substituting into the third.
407 Easy
407 of 555
Application +4 | -1
The equation of the bisectors of the angle between lines represented by equation 4x2−16xy−7y2=04 x^2-16 x y-7 y^2=0 is
View answer and explanation
Correct option: A
8x2+11xy−8y2=08 x^2+11 x y-8 y^2=0
Option A (8x2+11xy−8y2=08 x^2+11 x y-8 y^2=0) is correct because it is calculated using the formula for the equation of the bisectors of the angle between the lines.
408 Easy
408 of 555
Application +4 | -1
Let kk be an integer such that the triangle with vertices (k,−3k),(5,k)(k,-3 k),(5, k) and (−k,2)(-k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:
View answer and explanation
Correct option: D
(2,1/2)(2,1 / 2)
Option D (2,1/2)(2,1 / 2) is correct because it represents the orthocenter of the triangle, calculated using the given area and properties.
409 Easy
409 of 555
Analyzing +4 | -1
Two sides of an isosceles triangle are given by the equation 7x−y+3=07x - y + 3 = 0 and x+y−3=0.x + y - 3 = 0. If its third side passes through the point (1,-10), then its equations are
View answer and explanation
Correct option: C
x−3y−31=0x - 3y - 31 = 0 or, 3x+y+7=03x + y + 7 = 0
Option C (x−3y−31=0x - 3y - 31 = 0 or, 3x+y+7=03x + y + 7 = 0) is correct because these are the equations of the third side that pass through (1, -10) and form an isosceles triangle with the given sides.
410 Easy
410 of 555
Application +4 | -1
The vertices of a family of triangles have integer coordinates. If two of the vertices of all the triangles are (0, 0) and (6, 8), then the least value of areas of the triangles is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the minimum area of the triangle with integer coordinates is obtained when the third vertex is (1,1), resulting in an area of 1.
411 Easy
411 of 555
Application +4 | -1
The equation of the line passing through the points P(2,3)P(2,3) and Q(4,5)Q(4,5) is
View answer and explanation
Correct option: D
x−y+1=0x-y+1=0
Option D (x−y+1=0x-y+1=0) is correct because the equation of a line passing through two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) can be found using the determinant form or the two-point form, and substituting the given points yields the correct equation.
412 Medium
412 of 555
Application +4 | -1
The value of yy, so that the line through (3,y)(3, y) and (2,7)(2,7) is parallel to the line through (−1,4)(-1,4) and (0,6)(0,6), is
View answer and explanation
Correct option: C
9
Option C (9) is correct because parallel lines have equal slopes, so setting the slope between (3, y) and (2,7) equal to the slope between (-1,4) and (0,6) and solving for y gives y = 9.
413 Medium
413 of 555
Application +4 | -1
The area of triangle formed by the lines Y=axY=a x, x+y−a=0x+y-a=0 and the YY-axis is equal to
View answer and explanation
Correct option: C
12∣a1+a∣\frac{1}{2} | \frac{a}{1+a} |
Option D (a22∣1+a∣\frac{a^2}{2|1+a|}) is correct because it represents the area of the triangle formed by the given lines, calculated using the coordinates of the vertices.
414 Easy
414 of 555
Analyzing +4 | -1
Let ABCABC be a triangle, two of whose vertices are (15, 0) and (0, 10). If the orthocenter is (6, 9), then the third vertex is
View answer and explanation
Correct option: C
(0, 0)
Option C (0, 0) is correct because the third vertex satisfies the condition derived from the fact that the line passing through A and B is perpendicular to the line through C and the orthocenter O.
415 Easy
415 of 555
Application +4 | -1
The area of the parallelogram formed by the lines 2x−3y+a=0,3x−2y−a=0,2x−3y+3a=02 x-3 y+a=0, \quad 3 x-2 y-a=0,2 x-3 y+3 a=0 and 3x−2y−2a=03 x-2 y-2 a=0 in square units, is
View answer and explanation
Correct option: B
2a25\frac{2 a^2}{5}
Option B (2a25\frac{2 a^2}{5}) is correct because it is the area of the parallelogram calculated using the formula involving the constants and determinant of coefficients.
416 Medium
416 of 555
Analyzing +4 | -1
If θ\theta denotes the acute angle between the curves, y=10−x2 and y=2+x2y = 10 - {x^2}\,\mathrm{ and}\,\mathrm{ }y = 2 + {x^2} at a point of their intersection, then ∣tan⁡θ∣\left| {\tan \theta } \right| is equal to
View answer and explanation
Correct option: B
815\frac{8}{{15}}
Option B (8/15) is correct because it is the absolute value of the tangent of the acute angle between the two curves at their intersection points, calculated using their derivatives.
417 Easy
417 of 555
Analyzing +4 | -1
The number of points on the line x+y=4x + y = 4 which are unit distance apart from the line 2x+2y=52x + 2y = 5 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the distance between the two parallel lines is greater than 1, meaning no point on the first line is unit distance from the second.
418 Easy
418 of 555
Application +4 | -1
Lines 2x+y=12x + y = 1 and 2x+y=72x + y = 7 are
View answer and explanation
Correct option: A
On the same side of a point (0,12)\left( {0,\frac{1}{2}} \right)
The correct answer is A (On the same side of a point (0,1/2)) because when the point (0, 1/2) is substituted into both line equations, the resulting values have the same sign.
419 Easy
419 of 555
Application +4 | -1
Mirror image of (−3,5)(-3,5) in line mirror x−y+2=0x-y+2=0
View answer and explanation
Correct option: C
(3,−1)(3, -1)
Option C, (3, -1), is the correct mirror image of the point (-3, 5) with respect to the line x - y + 2 = 0.
420 Easy
420 of 555
Application +4 | -1
The area of the parallelogram formed by the lines xcos⁡α+ysin⁡α=p,xcos⁡α+ysin⁡α=qx \cos \alpha+y \sin \alpha=p, x \cos \alpha+y \sin \alpha=q, xcos⁡β+ysin⁡β=rx \cos \beta+y \sin \beta=r and xcos⁡β+ysin⁡β=sx \cos \beta+y \sin \beta=s, is
View answer and explanation
Correct option: C
∣(p−q)(r−s)sin⁡(α−β)∣\left|\frac{(p-q)(r-s)}{\sin (\alpha-\beta)}\right|
Option C is correct. The area of the parallelogram formed by the given lines is found using the formula involving the determinant of the coefficients and the difference of the constants, resulting in ∣(p−q)(r−s)sin⁡(α−β)∣\left|\frac{{(p-q)(r-s)}}{{\sin (\alpha -\beta )}}\right|.
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