1
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
2
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
$$y-2 x=5$$
3
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
$$3{x^2} + 5xy + 2{y^2} = 0$$
4
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
$$x^2-18x+80=0$$
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