1
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
If the origin is shifted to the point $\left(\frac{3}{2},-2\right)$ by the translation of axes, then the transformed equation of $2 x^2+4 x y+y^2+2 x-2 y+1=0$ is
A
$4 x^2+8 x y+2 y^2-16=0$
B
$2 x^2-4 x y+y^2=0$
C
$4 x^2+8 x y+2 y^2+9=0$
D
$2 x^2-4 x y+y^2+16=0$
2
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$L \equiv x \cos \alpha+y \sin \alpha-p=0$ represents a line perpendicular to the line $x+y+1=0$. If $p$ is positive, $\alpha$ lies in the fourth quadrant and perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line, $L=0$ is 5 units, then $p=$
A
5
B
$\frac{5}{2}$
C
10
D
$\frac{15}{2}$
3
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$(-2,-1),(2,5)$ are two vertices of a triangle and $\left(2, \frac{5}{3}\right)$ is its orthocenter. If $(m, n)$ is the third vertex of that triangle, then $m+n$ is equal to.
A
-4
B
-2
C
5
D
8
4
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$L_1 \equiv 2 x+y-3=0$ and $L_2 \equiv a x+b y+c=0$ are two equal sides of an isosceles triangle. If $L_3 \equiv x+2 y+1=0$ is the third side of this triangle and $(5,1)$ is a point on $L_2=$ 0 , then $\frac{b^2}{|a c|}=$
A
$\frac{121}{2}$
B
$\frac{49}{52}$
C
$\frac{81}{49}$
D
$\frac{25}{4}$
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