TS EAMCET 4-Aug-2021 Shift 2 Question Paper

‌

3

2

x+(2a−1)y=3. . . (i)
Slope m1=‌

−3∕2

2a−1

=‌

−3

2(2a−1)

‌2x+a2y=−3 . . . (ii)
=‌‌

−2

a2

∵ Lines are perpendicular.
∴‌‌‌

−3

2(2a−1)

×‌

(−2)

a2

=−1
⇒‌‌3m2=−1
⇒‌‌‌

3

a2(2a−1)

=−1
⇒‌‌2a3−a2+3=0
or ‌‌(a+1)(2a2−3a+3)=0
‌‌ or ‌‌a+1‌‌=0
⇒‌a‌‌=−1
∴‌‌ Lines are ‌

3

2

x−3y=3
⇒‌‌‌

x

2

−y=1⇒x−2y=2
2x+y=−3
On solving, we get
x=‌

−4

5

,y=‌

−7

5

Point of intersection (‌

−4

5

,‌

−7

5

)
Distance between (‌

−4

5

,‌

−7

5

) and (1,1)
‌‌=√(1+‌

4

5

)2+(1+‌

7

5

)2
‌‌=√(‌

9

5

)2+(‌

12

5

)2
‌‌=√‌

81+144

25

=‌

15

5

=3