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1
TS EAMCET 2022 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

The locus of the image of a variable point $(\alpha, 2 \alpha-1)$ with respect to the line $3 x-2 y+4=0$, is

A

$22(13 x+36)=19(13 y-11)$

B

$30(13 x+36)=19(13 y+37)$

C

$22(13 x+36)=7(13 y+11)$

D

$22(13 x-36)=30(13 y-11)$

2
TS EAMCET 2022 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $M$ be the foot of the perpendicular drawn from the point $(5,-7)$ to the line $3 x-5 y+1=0$. Then, the perpendicular distance from $M$ to the line $2 x+5 y-3=0$ is

A

$\frac{1}{2 \sqrt{29}}$

B

$\frac{9}{2 \sqrt{29}}$

C

$\frac{13}{2 \sqrt{29}}$

D

$\frac{3}{2 \sqrt{29}}$

3
TS EAMCET 2022 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $P$ is a point equidistant from all the vertices $A(-1,3), B(3,5), C(5,7)$ of a $\triangle A B C$, then $P A=$

A

11

B

$\sqrt{140}$

C

13

D

$\sqrt{130}$

4
TS EAMCET 2022 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

4 different pairs of lines are given in List I and the cosine of the angle between every pair of lines is given in List II. Match the following :

                                  List-I                     List-II
(A) 
<mspace width="1em"></mspace>
5 
<mi>x</mi>

<mn>2</mn>
+ 2 
<mn>7</mn>
x y − 
<mi>y</mi>

<mn>2</mn>
= 0 
<mspace width="1em"></mspace>
5 
<mi>x</mi>

<mn>2</mn>
+ 2 
<mn>7</mn>
x y − 
<mi>y</mi>

<mn>2</mn>
= 0 quad5x^(2)+2sqrt7xy-y^(2)=0		 (I) 
<msqrt>

  <mn>3</mn>

</msqrt>

<mn>2</mn>

<msqrt>

  <mn>3</mn>

</msqrt>

<mn>2</mn>
(sqrt3)/(2)
(B) 
<mspace width="1em"></mspace>

<mi>x</mi>

<mn>2</mn>
+ 
<mn>11</mn>
x y + 2 
<mi>y</mi>

<mn>2</mn>
= 0 
<mspace width="1em"></mspace>

<mi>x</mi>

<mn>2</mn>
+ 
<mn>11</mn>
x y + 2 
<mi>y</mi>

<mn>2</mn>
= 0 quadx^(2)+sqrt11xy+2y^(2)=0		 (II) 
<mo data-mjx-texclass="OPEN">(</mo>

<mfrac>

  <mn>1</mn>

  <mrow>

    <mn>2</mn>

    <msqrt>

      <mn>3</mn>

    </msqrt>

  </mrow>

</mfrac>

<mo data-mjx-texclass="CLOSE">)</mo>

<mrow>

  <mfrac>

    <mn>1</mn>

    <mrow>

      <mn>2</mn>

      <msqrt>

        <mn>3</mn>

      </msqrt>

    </mrow>

  </mfrac>  

</mrow>
((1)/(2sqrt3))
(C) 
<mspace width="1em"></mspace>

<mi>x</mi>

<mn>2</mn>
+ 2 
<mn>2</mn>
x y + 
<mi>y</mi>

<mn>2</mn>
= 0 
<mspace width="1em"></mspace>

<mi>x</mi>

<mn>2</mn>
+ 2 
<mn>2</mn>
x y + 
<mi>y</mi>

<mn>2</mn>
= 0 quadx^(2)+2sqrt2xy+y^(2)=0		 (III) 
<mn>1</mn>

<mn>2</mn>

<mn>1</mn>

<mn>2</mn>
(1)/(2)
(D) 
<mspace width="1em"></mspace>
3 
<mi>x</mi>

<mn>2</mn>
+ 4 
<mn>2</mn>
x y + 
<mi>y</mi>

<mn>2</mn>
= 0 
<mspace width="1em"></mspace>
3 
<mi>x</mi>

<mn>2</mn>
+ 4 
<mn>2</mn>
x y + 
<mi>y</mi>

<mn>2</mn>
= 0 quad3x^(2)+4sqrt2xy+y^(2)=0		 (IV) 
<mo data-mjx-texclass="OPEN">(</mo>

<mfrac>

  <mn>2</mn>

  <mn>3</mn>

</mfrac>

<mo data-mjx-texclass="CLOSE">)</mo>

<mrow>

  <mfrac>

    <mn>2</mn>

    <mn>3</mn>

  </mfrac>  

</mrow>
((2)/(3))
(V) 
<mn>1</mn>

<msqrt>

  <mn>2</mn>

</msqrt>

<mn>1</mn>

<msqrt>

  <mn>2</mn>

</msqrt>
(1)/(sqrt2)
$$ \text { The correct match is } $$
A
A B C D
III I V II
B
A B C D
III I IV V
C
A B C D
III I V IV
D
A B C D
III V II IV
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