TS EAMCET 2017 Code A Solved Paper

The point of intersection of two lines 5x−6y−1=0 and 3x+2y+5=0 is calculated as
(‌‌

b1c2−b2c1

a1b2−a2b1

,‌‌

c1a2−c2a1

a1b2−a2b1

) =(‌‌

−6×5−2×1

5×2−3×−6

,‌‌

−1×3−5×5

5×2−3×−6

)
=(‌‌

−30+2

10+18

,‌‌

−3−25

10+18

)
=(‌‌

−28

28

,‌‌

−28

28

)
=(−1,−1)
The slope of the line 3x−5y+11=0 is, m=‌‌

3

5

So, the equation of the line perpendicular to 3x−5y+11=0
and passing through (-1,-1) will be
(y+1)=−‌‌

5

3

(x+1)
3(y+1)=−5x−5
5x+3y+8=0