Find the equation of a line perpendicular to the line \(3 x+2 y+4=0\) and passing through the points \((5,6)\)
A. \(3 x+2 y-27=0\) B. \(2 y+x-3=\) 0 C. \(2 x-3 y+8=0\) D. \(3 x+2 y=-8\)
Correct Answer: C
Explanation
The required equation is given as: \(y-y_1=m\left(x-x_1\right)\) where \(m\) is the gradient which will be obtained from the given line \((3 x+2 y+4=0)\) for the line \(3 x+2 y+4=0\), \begin{array}{l} 2 y=-3 x-4 \\ y=\frac{-3 x}{2}-\frac{4}{2} \\ y=\frac{-3 x}{2}-2 \end{array} gradient \(=\frac{-3}{2}\) since the required line is perpendicular to the given line, then \(m \times\left(\frac{-3}{2}\right)=-1\) thus \(m=\frac{2}{3}\) therefore the required equation is \begin{array}{l} y-6=\frac{2}{3}(x-5) \\ =3 y-18=2 x-10 \\ 3 y-2 x-18+10=0 \\ 3 y-2 x-8=0 \end{array} if we multiply through by \(-1\), we get; \begin{array}{l} -3 y+2 x+8=0 \\ 2 x-3 y+8=0 \end{array}