TS EAMCET 6-Aug-2021 Shift 1 Question Paper

A B C D
2‌5‌4‌1
(A) x-intercept =5C [Let]
y-intercept =3C
Line: ‌

x

a

+‌

y

b

=1⇒‌

x

5C

+‌

y

3C

=1. . . (i)
Since, it passes through (−4,3).
Hence, −‌

4

5C

+‌

3

3C

=1⇒C=‌

1

5

From Eq. (i),
‌‌

x

5×‌ 1 5

+‌

y

3×‌ 1 5

=1⇒‌

x

1

+‌

5y

3

=1
⇒‌‌3x+5y=3
(B) Here given, that point P bisect the part intercepted between axes by line, that means P is mid-point of AB. Equation of line in one point form having slope =m is y+5‌‌=m(x−2)
y‌‌=mx−2m−5 . . . (i)
‌ Let ‌A‌‌=(0,−2m−5)⇒B=(2+‌

5

m

,0)
We know that, P is mid-point of A and B.
‌ Then, ‌2=‌

0+2+ 5 m

2

⇒‌‌4=2+‌

5

m

⇒m=‌

5

2

From Eq. (i),
y=‌

5

2

x−2(‌

5

2

)−5⇒5x−2y−20=0
(C) Any line parallel to 2x−3y+5=0 is
2x−3y+λ=0 . . . (i)
Since, Eq. (i) passes through (‌

2

5

,0),
Hence, 2×‌

2

5

−3×0+λ=0
⇒‌‌λ=−‌

4

5

From Eq. (i), 2x−3y−‌

4

5

=0
⇒10x−15y−4=0
(D) Any line perpendicular to 5x+2y+7=0 is
2x−5y+µ=0. . . (i)
Since, Eq. (i) passes through (0,‌

4

5

), then
2×0−5×‌

4

5

+µ=0⇒µ=4
From Eq. (i),
2x−5y+4=0