1
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1
Out of Syllabus

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
$$3{x^2} + 5xy + 2{y^2} = 0$$
2
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $$(\alpha,\beta)$$, then the quadratic equation whose roots are $$\alpha+4\beta$$ and $$4\alpha+\beta$$, is :

A
$$x^2-20x+99=0$$
B
$$x^2-22x+120=0$$
C
$$x^2-19x+90=0$$
D
$$x^2-18x+80=0$$
3
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

Let $$B$$ and $$C$$ be the two points on the line $$y+x=0$$ such that $$B$$ and $$C$$ are symmetric with respect to the origin. Suppose $$A$$ is a point on $$y-2 x=2$$ such that $$\triangle A B C$$ is an equilateral triangle. Then, the area of the $$\triangle A B C$$ is :

A
$$\frac{10}{\sqrt{3}}$$
B
$$2 \sqrt{3}$$
C
$$3 \sqrt{3}$$
D
$$\frac{8}{\sqrt{3}}$$
4
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

A light ray emits from the origin making an angle 30$$^\circ$$ with the positive $$x$$-axis. After getting reflected by the line $$x+y=1$$, if this ray intersects $$x$$-axis at Q, then the abscissa of Q is :

A
$${2 \over {\left( {\sqrt 3 - 1} \right)}}$$
B
$${2 \over {3 - \sqrt 3 }}$$
C
$${{\sqrt 3 } \over {2\left( {\sqrt 3 + 1} \right)}}$$
D
$${2 \over {3 + \sqrt 3 }}$$
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