Find the equation of the line parallel to `x-2y+3=0` and passing through the point (1,-2).

Explanation

Given line `x-2 y+3=0` can be written as `y=1/12 x+3/2`

Slope of the line `(1)` is `m_1=1/2`.
Therefore, slope of the line perpendicular to line `(1)` is `m_2=-1/m_1=-2`

Equation of the line with slope `-2` and passing through the point `(1,-2)` is
`y-(-2)=-2(x-1)`or `y=-2 x`. which is the required equation.