TS ICET 23 May 2018 Shift 1 Paper

The point of intersection of the lines 3x+4y+5=0 and 4x+y–2=0 is obtained by solving the two equations and we get
x=1 and y=–2
Hence the point of intersection is (1, –2)
Slope of the line

x

5

+

y

2

=1 is –2/5

x

5

+

y

2

=1 is

5

2

Required equation of a line passing through (1,−2) and having a slope of 5/2 is
y−(−2)=(

5

2

)(x−1) i.e. 2y+4=5x−5
i.e 5x−2y−9=0