Respuesta :
Given:
The point A = (2,-3)
The equation given is 2x +5y =4
We will obtain the slope (m1) from the equation of the line since the line with point (2,-3) is perpendicular to the equation.
The rule for perpendicularism is,
[tex]m_1m_2=-1[/tex]Isolate y from the equation
[tex]\begin{gathered} 2x+5y=4 \\ 5y=4-2x \\ \frac{5y}{5}=-\frac{2x}{5}+\frac{4}{5} \\ y=-\frac{2}{5}x+\frac{4}{5} \end{gathered}[/tex]The equation of a line is written in the form
[tex]y=mx+b[/tex]Therefore,
[tex]m_2=-\frac{2}{5}[/tex]Let us now obtain m1
[tex]\begin{gathered} m_1(-\frac{2}{5})=-1 \\ -\frac{2}{5}m_1=-1 \\ -2m_1=-1\times5 \\ -2m_1=-5 \\ m_1=\frac{-5}{-2}=\frac{5}{2} \\ \therefore m_1=\frac{5}{2} \end{gathered}[/tex]The formula for the equation of a line given a point is,
[tex]y-y_1=m(x-x_1)[/tex]Where,
[tex](x_1,y_1)=(2,-3)[/tex]Therefore,
[tex]\begin{gathered} y-(-3)=\frac{5}{2}(x-2) \\ y+3=\frac{5}{2}\times x-\frac{5}{2}\times2 \\ y+3=\frac{5}{2}x-5 \\ y=\frac{5}{2}x-5-3 \\ \therefore y=\frac{5}{2}x-8 \end{gathered}[/tex]Hence, the equation of the line is,
[tex]y=\frac{5}{2}x-8[/tex]
