1
JEE Main 2023 (Online) 13th April Evening Shift
+4
-1 Let $$(\alpha, \beta)$$ be the centroid of the triangle formed by the lines $$15 x-y=82,6 x-5 y=-4$$ and $$9 x+4 y=17$$. Then $$\alpha+2 \beta$$ and $$2 \alpha-\beta$$ are the roots of the equation :

A
$$x^{2}-7 x+12=0$$
B
$$x^{2}-13 x+42=0$$
C
$$x^{2}-14 x+48=0$$
D
$$x^{2}-10 x+25=0$$
2
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$$\sqrt3$$
3
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
4
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
$$y-2 x=5$$
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