$A B C$ is a right angled isosceles triangle, right angled at $A(2,1)$. If the equation of side $B C$ is $-2 x+y=3$, then the combined equation of lines $A B$ and $A C$ is

$3 x^2-3 y^2+8 x y+20 x+10 y+25=0$

$3 x^2-3 y^2+8 x y-20 x-10 y+25=0$

$3 x^2-3 y^2+8 x y+10 x+15 y+20=0$

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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Medium

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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.

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Showing 1-30 of 139 questions

1 Medium

1 of 139

Application +4 | -1

The orthocentre of the triangle whose vertices are

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

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the orthocenter of the triangle formed by the given vertices on a parabola.

2 Medium

2 of 139

Application +4 | -1

(a+3 b)(3 a+b)=4 h^2

, then the angle between the lines represented by

a x^2+2 h x y+b y^2=0

View answer and explanation

Correct option: C

60^{\circ}

Option C (

60^{\circ}

) is correct because it is calculated using the given condition

(a+3b)(3a+b)=4h^2

and the formula for the angle between the lines.

3 Medium

3 of 139

Application +4 | -1

The point

\left(a^2, a+1\right)

lies in the angle between the lines

3 x-y+1=0

and

x+2 y-5=0

containing the origin, if

View answer and explanation

Correct option: A

a \in(-3,0) \cup\left(\frac{1}{3}, 1\right)

Option A (

a \in(-3,0) \cup(\frac{1}{3}, 1)

) is correct because the point and the origin must lie on the same side of both lines, leading to two inequalities that define the range of 'a'.

4 Medium

4 of 139

Application +4 | -1

The incentre of the triangle formed by (0, 0), (5, 12), (16, 12) is

View answer and explanation

Correct option: A

(7, 9)

Option A (7, 9) is correct because the incenter is calculated using the formula

\left( {\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\;\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)

, where

a, b, c

are the side lengths of the triangle.

5 Medium

5 of 139

Application +4 | -1

The equation of the bisectors of angle between the lines represented by equation

(y-m x)^2=(x+m y)^2

View answer and explanation

Correct option: A

m x^2+\left(m^2-1\right) x y-m y^2=0

Option A (

m x^2+\left(m^2-1\right) x y-m y^2=0

) is correct because it is calculated using the formula for the equation of the bisectors of the angle between the lines.

6 Medium

6 of 139

Application +4 | -1

The reflection of the curve

x y=1

in the line

y=2 x

is the curve

12 x^2+r x y+s y^2+t=0

, then the value of

r

is-

View answer and explanation

Correct option: A

-7

Option A (-7) is correct because it is the coefficient of the xy term in the reflected curve equation, obtained by finding the reflection of a point on the original curve and substituting it back into the equation.

7 Medium

7 of 139

Application +4 | -1

The line

3 x+2 y=24

meets the

y

-axis at

A

and the

x

-axis at

B ; C

is a point on the perpendicular bisector of

A B

such that the area of the triangle

A B C

91 \mathrm{sq}

. units. The coordinates of

C

can be

View answer and explanation

Correct option: B

(29 / 2,13)

Option B ((29 / 2, 13)) is correct because it satisfies both the perpendicular bisector equation and the area condition for triangle ABC.

8 Medium

8 of 139

Analyzing +4 | -1

{t_1},{t_2}

and

{t_3}

are distinct points

\left( {{t_1},2a{t_1} + at_1^3} \right),\left( {{t_2},2a{t_2} + at_2^3} \right)

and

{t_3},2a{t_3} + at_3^3)

are collinear, if

View answer and explanation

Correct option: C

{t_1} + {t_2} + {t_3} = 0

Option C (

{t_1} + {t_2} + {t_3} = 0

) is correct because the collinearity condition of the three points leads to this relationship between

t_1, t_2,

and

t_3

9 Medium

9 of 139

Analyzing +4 | -1

The equation

\lambda\left(x^3-3 x y^2\right)+y^3-3 x^2 y=0

represents

View answer and explanation

Correct option: A

three lines equally inclined to the other

Option A (three lines equally inclined to the other) is correct because the equation represents three lines passing through the origin, equally inclined to each other.

10 Medium

10 of 139

Application +4 | -1

{P}

lies on the line

8 {y}-15 {x}=0

and

{Q}

lies on the line

10 {y}-3 {x}=0

and the mid-point of

PQ

(8,6)

. The distance

P Q

is-

View answer and explanation

Correct option: B

\frac{60}{7}

Option B (

\frac{60}{7}

) is correct because it represents the distance PQ calculated by finding the coordinates of P and Q using the given information about the midpoint and the lines they lie on.

11 Medium

11 of 139

Application +4 | -1

If the sum of the slopes of the lines represented by the equation

x^2-2 x y \tan A-y^2=0

be 4, then

\angle A=

View answer and explanation

Correct option: D

\tan ^{-1}(-2)

Option D (

\tan ^{-1}(-2)

) is correct because it is derived from the relationship between the sum of slopes and the coefficient of the xy term in the given equation.

12 Medium

12 of 139

Application +4 | -1

The equation of a straight line passing through the point of intersection of

x - y + 1 = 0

and

3x + y - 5 = 0

and perpendicular to one of them, is

View answer and explanation

Correct option: D

x - 3y + 5 = 0

Option D (

x - 3y + 5 = 0

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

13 Medium

13 of 139

Application +4 | -1

The equation to the line which passes through the point of intersection of the two lines

2 x+3 y-1=0

and

3 x+2 y+1=0

, and is normal to the line joining

(2,4),(4,7)

is-

View answer and explanation

Correct option: D

None of these

Option A (

2 x+3 y=18

) is correct because it is the equation of the perpendicular bisector of the line segment AB, found by calculating the midpoint and slope of AB and then using the point-slope form.

14 Medium

14 of 139

Application +4 | -1

3 a-2 b+5 c=0

, then the family of the lines

a x+b y+c=0

an always concurrent a point whole coordinates are....

View answer and explanation

Correct option: B

\left(\frac{3}{5}, \frac{-2}{5}\right)

Option B (

\left(\frac{3}{5}, \frac{{-2}}{5}\right)

) is correct because it satisfies the equation derived from the given condition

3a - 2b + 5c = 0

15 Medium

15 of 139

Application +4 | -1

The points

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

and

\left( {13,\lambda } \right)

lie on the same straight line, if

\lambda

is equal to

View answer and explanation

Correct option: B

- 7

Option B (-7) is correct because the collinearity condition (zero area of the triangle formed by the points) leads to λ = -7.

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16 Medium

16 of 139

Application +4 | -1

The pair of points which lie on the same side the straight line

3 x-8 y-7=0

, is

View answer and explanation

Correct option: C

(-1,-1),(3,7)

The pair of points (-1,-1) and (3,7) lie on the same side of the line

3x - 8y - 7 = 0

because when substituted into the equation, both points result in values with the same sign.

17 Medium

17 of 139

Analyzing +4 | -1

Consider the set of all lines

px + qy + r = 0

such that

3p + 2q + 4r = 0

Which one of the following statements is true?

View answer and explanation

Correct option: B

The lines are concurrent at the point

\left( {\frac{3}{4},\frac{1}{2}} \right)

Option B (The lines are concurrent at the point (3/4, 1/2)) is correct because all lines satisfying the given condition pass through the point (3/4, 1/2).

18 Medium

18 of 139

Application +4 | -1

The measure of the angle between lines

x+y=0

and

y=[\pi]

=\ldots .

View answer and explanation

Correct option: A

\frac{\pi}{4}

Option A (

\frac{\pi}{4}

) is correct because the angle between the lines is found using the slopes of the lines and the tangent formula.

19 Medium

19 of 139

Analyzing +4 | -1

If the point

(1+\cos \theta, \sin \theta)

lies between the region corresponding to the acute angle between the lines

3 y=x

and

6 y=x

, then

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the solution requires finding the range of θ that satisfies the inequality derived from the point lying between the given lines, which doesn't match options A, B, or C.

20 Medium

20 of 139

Application +4 | -1

The condition that the pair of straight line joining the origin to the intersections of the line

y=m x+c

and the circle

x^2+y^2=a^2

is at the right angles, is 

View answer and explanation

Correct option: B

2 c^2=a^2\left(1+m^2\right)

Option B (

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.

21 Medium

21 of 139

Application +4 | -1

y-intercept of line passes through (2, 2) and is perpendicular to the line

3x + y = 3

, is

View answer and explanation

Correct option: D

\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.

22 Medium

22 of 139

Application +4 | -1

The equation of the obtuse angle bisector of lines

12 x-5 y+7=0

and

3 y-4 x-1=0

View answer and explanation

Correct option: A

4 x+7 y+11=0

Option A (

4 x+7 y+11=0

) is correct because it is the equation of the obtuse angle bisector, determined by checking the sign of

a_1a_2 + b_1b_2

23 Medium

23 of 139

Analyzing +4 | -1

A variable straight line drawn through the point of intersection of lines

\frac{x}{a}+\frac{y}{b}=1

and

\frac{x}{b}+\frac{y}{a}=1

, meets the coordinate axes in

A

and

B

. Then, the locus of the mid-point of

A B

View answer and explanation

Correct option: D

2 x y(a+b)=a b(x+y)

Option D (

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.

24 Medium

24 of 139

Analyzing +4 | -1

If the straight lines

a_1 x+b_1 y+c_1=0

a_1 x+b_1 y+c_2=0, a_2 x+b_2 y+d_1=0

and

a_2 x+b_2 y+d_2=0

are the sides of rhombus, then

View answer and explanation

Correct option: A

\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.

25 Medium

25 of 139

Application +4 | -1

Let P=(-1,0),Q=(0,0) and

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

\sqrt 3 x + y = 0

Option A (

\sqrt 3 x + y = 0

) is correct because the bisector's equation is derived from the angle between the lines and the slope.

26 Medium

26 of 139

Application +4 | -1

Area of the triangle formed by the lines

y = 2x,y = 3x

and

y = 5

is equal to (in square unit)

View answer and explanation

Correct option: B

\frac{{25}}{{12}}

Option B

\frac{{25}}{{12}}

is correct because it is the area of the triangle calculated using the coordinates of the intersection points.

27 Medium

27 of 139

Application +4 | -1

3 a-2 b+5 c=0

, family of the lines

a x+b y+c=0

are always concurrent at a point whose coordinates are

View answer and explanation

Correct option: B

\left(\frac{3}{5},-\frac{2}{5}\right)

Option B (

\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 = 0

28 Medium

28 of 139

Application +4 | -1

The lines

x+2 y+3=0, x+2 y-7=0

and

2 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

2 x-y-14=0

Option A (

2x-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.

29 Medium

29 of 139

Analyzing +4 | -1

A B C

is a right angled isosceles triangle, right angled at

A(2,1)

. If the equation of side

B C

-2 x+y=3

, then the combined equation of lines

A B

and

A C

View answer and explanation

Correct option: D

None of the above

The correct answer is None of the above because the combined equation of lines AB and AC is

3x^2 - 8xy - 3y^2 - 4x + 22y - 7 = 0

, which is not among the options.

30 Medium

30 of 139

Analyzing +4 | -1

A B C D

is square

A \equiv(1,2), B \equiv(3,-4)

. If line

C D

passes through

(3,8)

, then mid-point of

C D

View answer and explanation

Correct option: D

\left(\frac{28}{5}, \frac{1}{5}\right)

Option D (

\left(\frac{{28}}{{5}}, \frac{{1}}{{5}}\right)

) is correct because it is the midpoint of CD, found by solving the equations of line CD and the perpendicular bisector of AB.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 139 single-correct-answer MCQs on STRAIGHT LINES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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