Question

find the equation of a line perpendicular to the line x- 2y +3 = 0 and passing through the points (1,-2)

Answer 1

$$\begin{gathered} \mathrm{Slop}\mathrm{of}\mathrm{line}\mathrm{x}-2\mathrm{y}+3=0\mathrm{is}=\frac{1}{2} \\ \mathrm{So}\mathrm{slop}\mathrm{of}\mathrm{line}\mathrm{perpendicular}\mathrm{to}\mathrm{it}=\frac{-1}{{\displaystyle \frac{1}{2}}}=-2 \\ \mathrm{So}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{pasing}\mathrm{from}\left(1,-2\right)\mathrm{and}\mathrm{having}\mathrm{slop}-2\mathrm{is} \\ \mathrm{y}+2=-2\left(\mathrm{x}-1\right) \\ 2\mathrm{x}+\mathrm{y}=0 \end{gathered}$$