Find the equation of a line perpendicular to | vedclass

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(A) The given line is $x-2y+3=0$,which can be rewritten as $2y = x+3$,or $y = \frac{1}{2}x + \frac{3}{2}$.The slope of this line is $m_1 = \frac{1}{2}

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$2x+y=0$

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(A) The given line is $x-2y+3=0$,which can be rewritten as $2y = x+3$,or $y = \frac{1}{2}x + \frac{3}{2}$.The slope of this line is $m_1 = \frac{1}{2}$.The slope of a line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2$.The equation of the line passing through the point $(1, -2)$ with slope $m_2 = -2$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.Substituting the values: $y - (-2) = -2(x - 1)$.$y + 2 = -2x + 2$.$2x + y = 0$.

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

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