TS ICET 23 May 2018 Shift 1 Paper

The point of intersection of the lines $3\mathit{x}+4\mathit{y}+5=0$ and $4\mathit{x}+\mathit{y}\mathit{–}2=0$ is obtained by solving the two equations and we get
$\mathit{x}=1$ and $\mathit{y}=\mathit{–}2$
Hence the point of intersection is (1, –2)
Slope of the line $\frac{\mathit{x}}{5}+\frac{\mathit{y}}{2}=1$ is –2/5
$\frac{\mathit{x}}{5}+\frac{\mathit{y}}{2}=1$ is $\frac{5}{2}$
Required equation of a line passing through $\left(1,-2\right)$ and having a slope of 5/2 is
$\mathit{y}-\left(-2\right)=\left(\frac{5}{2}\right)\left(\mathit{x}-1\right)$ i.e. $2\mathit{y}+4=5\mathit{x}-5$
i.e $5\mathit{x}-2\mathit{y}-9=0$