1
JEE Main 2023 (Online) 12th April Morning Shift
+4
-1 If the point $$\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$$ lies on the curve traced by the mid-points of the line segments of the lines $$x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$$ between the co-ordinates axes, then $$\alpha$$ is equal to :

A
$$-$$7
B
7
C
$$-$$7$$\sqrt3$$
D
7$$\sqrt3$$
2
JEE Main 2023 (Online) 8th April Morning Shift
+4
-1 Let $$C(\alpha, \beta)$$ be the circumcenter of the triangle formed by the lines

$$4 x+3 y=69$$

$$4 y-3 x=17$$, and

$$x+7 y=61$$.

Then $$(\alpha-\beta)^{2}+\alpha+\beta$$ is equal to :

A
15
B
17
C
16
D
18
3
JEE Main 2023 (Online) 6th April Morning Shift
+4
-1 The straight lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$ pass through the origin and trisect the line segment of the line L: $$9 x+5 y=45$$ between the axes. If $$\mathrm{m}_{1}$$ and $$\mathrm{m}_{2}$$ are the slopes of the lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$, then the point of intersection of the line $$\mathrm{y=\left(m_{1}+m_{2}\right)}x$$ with L lies on :

A
$$6 x-y=15$$
B
$$6 x+y=10$$
C
$$\mathrm{y}-x=5$$
D
$$y-2 x=5$$
4
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1 The combined equation of the two lines $$ax+by+c=0$$ and $$a'x+b'y+c'=0$$ can be written as

$$(ax+by+c)(a'x+b'y+c')=0$$.

The equation of the angle bisectors of the lines represented by the equation $$2x^2+xy-3y^2=0$$ is :

A
$$3{x^2} + xy - 2{y^2} = 0$$
B
$${x^2} - {y^2} - 10xy = 0$$
C
$${x^2} - {y^2} + 10xy = 0$$
D
$$3{x^2} + 5xy + 2{y^2} = 0$$
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination