The line passing through $\left( { - 1,\frac{{\rm{\pi }}}{2}} \right)$ and perpendicular to $\sqrt 3 \sin {\rm{\theta }} + 2\cos {\rm{\theta }} = \frac{4}{r}$ is

$2 = \sqrt 3 r\cos {\rm{\theta }} - 2r\sin {\rm{\theta }}$

$5 = - 2\sqrt 3 r\sin {\rm{\theta }} + 4r\cos {\rm{\theta }}$

$2 = \sqrt 3 r\cos {\rm{\theta }} + 2r\sin {\rm{\theta }}$

$5 = 2\sqrt 3 r\sin {\rm{\theta }} + 4r\cos {\rm{\theta }}$

The image of $P(a, b)$ on $y=-x$ is $Q$ and the image of $Q$ on the line $y=x$ is $R$. Then the mid point of $R$ is

$\left(\frac{a+b}{2}, \frac{b+a}{2}\right)$

The equation of a straight line which passes through the point $\left( {a\;{\rm{co}}{{\rm{s}}^3}{\rm{\theta }},{\rm{\;}}a\;{\rm{si}}{{\rm{n}}^3}{\rm{\theta }}} \right)$ and perpendicular to $x\sec {\rm{\theta }} + y\;{\rm{cosec\;\theta }} = a$ is

$x\cos {\rm{\theta }} - \;y\sin {\rm{\theta }} = - a\cos 2{\rm{\theta }}$

$\frac{x}{a} + \frac{y}{a} = a\cos {\rm{\theta }}$

$x\cos {\rm{\theta }} + y\sin {\rm{\theta }} = a\cos 2{\rm{\theta }}$

$x\cos {\rm{\theta }} + y\sin {\rm{\theta }} - a\cos 2{\rm{\theta }}$

Back to STRAIGHT LINES

Your Progress

0 / 408 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 211-240 of 408 questions

Preparing practice and saved progress…

211 Easy

211 of 408

Analyzing +4 | -1

The foot of the perpendicular drawn from the origin, on the line,

3 x+y=\lambda(\lambda \neq 0)

P

. If the line meets

x

-axis at

A

and

y

-axis at

B

, then the ratio

B P: P A

View answer and explanation

Correct option: D

3: 1

Option D (

3: 1

) is correct because it represents the ratio of BP to PA, calculated using the coordinates of the points and the distance formula.

212 Easy

212 of 408

Application +4 | -1

a, b, c

are in A.P., then

a x+b y+c=0

represents

View answer and explanation

Correct option: B

a family of concurrent lines

Option B (a family of concurrent lines) is correct because the given equation can be rewritten to represent lines passing through the intersection of two other lines.

213 Easy

213 of 408

Application +4 | -1

The distance between the parallel lines

y = x + a,\;y = x + b

View answer and explanation

Correct option: A

\frac{{\left| {b - a} \right|}}{{\sqrt 2 }}

The correct answer is A (|b - a|/√2) because the distance between two parallel lines is calculated using the formula.

214 Easy

214 of 408

Application +4 | -1

If the points

\left( { - 2,\; - 5} \right),\left( {2,\; - 2} \right),\;\left( {8,\;a} \right)

are collinear, then the value of

a

View answer and explanation

Correct option: B

\frac{5}{2}

Option B (

\frac{5}{2}

) is correct because the slope between any two points on a line must be equal if the points are collinear.

215 Easy

215 of 408

Application +4 | -1

A straight line through P(1,2) is such that its intercept between the axes is bisected at P. Its equation is

View answer and explanation

Correct option: D

2x + y = 4

Option D (

2x + y = 4

) is correct because it represents the equation of the line whose intercept between the axes is bisected at P(1,2).

216 Easy

216 of 408

Application +4 | -1

Circumcentre of triangle whose vertices are (0, 0), (3, 0) and (0, 4) is

View answer and explanation

Correct option: A

\left( {\frac{3}{2},2} \right)

Option A

\left( {\frac{{3}}{2},2} \right)

is correct because the circumcentre of a right-angled triangle is the midpoint of the hypotenuse.

217 Easy

217 of 408

Application +4 | -1

Let the perpendiculars from any point on the line

2x + 11y = 5

upon the lines

24x + 7y - 20 = 0

and

4x - 3y - 2 = 0

have the lengths

{p_1}

and

{p_2}

respectively. Then,

View answer and explanation

Correct option: B

{p_1} = {p_2}

Option B (

{p_1} = {p_2}

) is correct because the line

2x + 11y = 5

is the angle bisector of the two lines, hence the perpendicular distances are equal.

218 Easy

218 of 408

Application +4 | -1

The base vertices of an isosceles triangle PQR are

Q(1,3)

and

R(-2,7)

. The vertex

P

can be

View answer and explanation

Correct option: C

(5 / 6,6)

Option C (

(5 / 6,6)

) is correct because it satisfies the equation of the perpendicular bisector of the line segment QR, which is the locus of point P.

219 Easy

219 of 408

Analyzing +4 | -1

The equation of line through the point (1, 2) whose distance from the point (3, 1) has the greatest value, is

View answer and explanation

Correct option: C

x + 2y = 5

Option C (

x + 2y = 5

) is correct because the line that maximizes the distance between two points passes through both points.

220 Easy

220 of 408

Application +4 | -1

The angle between the two straight lines

2 x^2-5 x y+2 y^2-3 x+3 y+1=0

View answer and explanation

Correct option: D

\tan ^{-1} \frac{3}{4}

Option D (

\tan ^{-1} \frac{3}{4}

) is correct because it is calculated using the formula for the angle between the two straight lines represented by the given equation.

221 Easy

221 of 408

Application +4 | -1

What is the equation of the locus of a point which moves such that 4 times its distance from the

x

-axis is the square of its distance from the origin?

View answer and explanation

Correct option: B

{x^2} + {y^2} - 4\left| y \right| = 0

Option B (

{x^2} + {y^2} - 4\left| y \right| = 0

) is correct because it directly represents the locus of a point where 4 times its distance from the x-axis equals the square of its distance from the origin.

222 Easy

222 of 408

Analyzing +4 | -1

If two vertices of a triangle are

\left( { - 2,3} \right)

and

\left( {5, - 1} \right)

. Orthocentre lies at the origin and centroid on the line

x + y = 7

, then the third vertex lies at

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the calculated third vertex does not match any of the given options after applying the orthocenter and centroid conditions.

223 Easy

223 of 408

Application +4 | -1

If one vertex of an equilateral triangle is at (2,-1) and the base is

x + y - 2 = 0,

then the length of each side is

View answer and explanation

Correct option: B

\sqrt {2/3}

Option B (

\sqrt {2/3}

) is correct because it is the side length of the equilateral triangle calculated using the perpendicular distance from the vertex to the base.

224 Easy

224 of 408

Application +4 | -1

The line passing through

\left( { - 1,\frac{\mathrm{\pi }}{2}} \right)

and perpendicular to

\sqrt 3 \sin \mathrm{\theta } + 2\cos \mathrm{\theta } = \frac{4}{r}

View answer and explanation

Correct option: A

2 = \sqrt 3 r\cos \mathrm{\theta } - 2r\sin \mathrm{\theta }

Option A (

2 = \sqrt 3 r\cos \mathrm{\theta } - 2r\sin \mathrm{\theta }

) is correct because it represents the line perpendicular to the given line and passing through the specified point in polar coordinates.

225 Easy

225 of 408

Application +4 | -1

The image of

P(a, b)

y=-x

Q

and the image of

Q

on the line

y=x

R

. Then the mid point of

R

View answer and explanation

Correct option: D

(0,0)

Option D ((0,0)) is correct because the image of P(a, b) on y = -x is Q(-b, -a), and the image of Q on y = x is R(-a, -b). The midpoint of PR is then ((a - a)/2, (b - b)/2) = (0, 0).

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

226 Easy

226 of 408

Application +4 | -1

If three points

(h, 0),(a, b)

and

(o, k)

lie on a line then

View answer and explanation

Correct option: B

\frac{a}{h}+\frac{b}{k}=1

Option B (

\frac{a}{h}+\frac{b}{k}=1

) is correct because it is derived from the condition that the determinant of the matrix formed by the points is zero, indicating collinearity.

227 Easy

227 of 408

Application +4 | -1

Points on the line

x + y = 4

that lie at a unit distance from the line

4x + 3y - 10 = 0,

are

View answer and explanation

Correct option: A

(3,1) and (-7,11)

Option A ((3,1) and (-7,11)) is correct because these are the points on the line x+y=4 that are at a unit distance from the line 4x+3y-10=0.

228 Easy

228 of 408

Application +4 | -1

The equation of a straight line which passes through the point

\left( {a\;\mathrm{co}{\mathrm{s}^3}\mathrm{\theta },\mathrm{\;}a\;\mathrm{si}{\mathrm{n}^3}\mathrm{\theta }} \right)

and perpendicular to

x\sec \mathrm{\theta } + y\;\mathrm{cosec\;\theta } = a

View answer and explanation

Correct option: B

x\cos \mathrm{\theta } - \;y\sin \mathrm{\theta } = - a\cos 2\mathrm{\theta }

Option B (

x\cos \mathrm{\theta } - \;y\sin \mathrm{\theta } = - a\cos 2\mathrm{\theta }

) is correct because it represents the equation of a line passing through the given point and perpendicular to the given line.

229 Easy

229 of 408

Application +4 | -1

Equation of the straight line making equal intercepts on the axes and passing through the point (2, 4), is

View answer and explanation

Correct option: C

x + y - 6 = 0

The correct answer is C (x + y - 6 = 0) because the equation of a line making equal intercepts on the axes and passing through (2, 4) is found.

230 Easy

230 of 408

Analyzing +4 | -1

{t_1} + {t_2} + {t_3} = - {t_1}{t_2}{t_3}

, then orthocentre of the triangle formed by the points

\left[ {a{t_1}{t_2},a\left( {{t_1} + {t_2}} \right)} \right],\left[ {a{t_2}{t_3},a\left( {{t_2} + {t_3}} \right)} \right]

and

\left[ {a{t_3}{t_1},a\left( {{t_3} + {t_1}} \right)} \right]

, lies on

View answer and explanation

Correct option: B

\left( { - a,\;0} \right)

Option B ((-a, 0)) is correct because the orthocenter of the triangle formed by the given points lies on the line x = -a, which means the orthocenter is (-a, 0).

231 Easy

231 of 408

Application +4 | -1

Equation of a line passing through the line of interception of lines

2x - 3y + 4 = 0,\;

3x + 4y - 5 = 0

and perpendicular to

6x - 7y + 3 = 0

View answer and explanation

Correct option: B

119x + 102y = 125

Option B (

119x + 102y = 125

) is correct because it represents a line passing through the intersection of the given lines and is perpendicular to the specified line.

232 Easy

232 of 408

Application +4 | -1

A point moves in such a way that the square of its distance from the point (3,-2) is equal to numerically its distance from the line

5x - 12y = 13

. The equation of the locus of the point is

View answer and explanation

Correct option: C

13\left( {{x^2} + {y^2}} \right) - 83x + 64y + 182 = 0

Option C (

13\left( {{x^2} + {y^2}} \right) - 83x + 64y + 182 = 0

) is correct because it represents the locus of the point satisfying the given condition.

233 Easy

233 of 408

Application +4 | -1

{x_1},\;{x_2},\;{x_3}

as well as

{y_1},\;{y_2},{y_3}\;

are in GP with the same common ratio, then the points

\left( {{x_1},\;{y_1}} \right),\left( {{x_2},\;{y_2}} \right)

and

\left( {{x_3},\;{y_3}} \right)

View answer and explanation

Correct option: D

Lie on a straight line

Option D (Lie on a straight line) is correct because the ratio of the y-coordinates to the x-coordinates is constant, indicating collinearity.

234 Easy

234 of 408

Application +4 | -1

Two points A and B move on the coordinate axes such that the distance between them remains same. The locus of the mid-point of AB is

View answer and explanation

Correct option: C

A circle

Option C (A circle) is correct because the locus of the midpoint of AB, where A and B move on the coordinate axes such that the distance between them remains constant, is a circle.

235 Easy

235 of 408

Application +4 | -1

If 3, 4 are intercepts of a line L≡0, then the distance of L≡0 from the origin is

View answer and explanation

Correct option: D

\frac{{12}}{5}

units

Option D (

x - 7y + 6 = 0

) is correct because it represents the other line when 2x-y=1 bisects the angle between it and y=x.

236 Easy

236 of 408

Application +4 | -1

The point

(2,1)

is shifted by

3 \sqrt{2}

units parallel to the line

x+y=1

, in the direction of increasing ordinate, to reach

Q

. The image of

Q

in the line

x+y=1

, is

View answer and explanation

Correct option: B

(-3,2)

Option B ((-3,2)) is correct because it represents the image of the translated point Q in the line x+y=1.

237 Easy

237 of 408

Application +4 | -1

The equation of a line passing through

\left(a \cos ^3 \theta, a \sin ^3 \theta\right)

and perpendicular to the line

x \sec \theta+y \operatorname{cosec} \theta=a

, is

View answer and explanation

Correct option: B

x \cos \theta-y \sin \theta=a \cos 2 \theta

Option B (

x \cos \theta-y \sin \theta=a \cos 2 \theta

) is correct because it represents the equation of the line passing through the given point and perpendicular to the given line.

238 Easy

238 of 408

Application +4 | -1

In a triangle

A B C

, coordinates of

A

are

(1,2)

and the equations of medians through

B

and

C

are

x+y=5

and

x=4

respectively. The coordinates of

C

and

B

are respectively

View answer and explanation

Correct option: D

(4,3),(7,-2)

Option D ((4,3),(7,-2)) is correct because it satisfies the median equations and the coordinates of vertex A.

239 Easy

239 of 408

Application +4 | -1

Orthocentre of the triangle whose sides are given by

4\;x - 7\;y + 10 = 0,x + y - 5 = 0

and

7\;x + 4\;y - 15 = 0

View answer and explanation

Correct option: D

\left( {1,2} \right)

Option D ((1,2)) is correct because the intersection of two perpendicular sides gives the orthocenter.

240 Easy

240 of 408

Analyzing +4 | -1

The equation

\sqrt{\left(x^2+4 y^2-4 x y+4\right)}+x-2 y=1

represents a/an

View answer and explanation

Correct option: A

straight line

The correct answer is A (straight line) because after squaring and simplifying, the given equation reduces to a linear equation.

Revision notes and key takeaways

Review STRAIGHT LINES in your textbook or class notes, then use this set to check your understanding.

Showing 211-240 of 408 questions

Your Progress

Straight lines practice · Easy

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice