How to solve for "Find the equation of the line that is perpendicular to y = -2/3x and contains point (4,-8)" ?

Answer:
[tex]y=\frac{3}{2}x+(-14)[/tex]
Step-by-step explanation:
Equation of a line has been given as,
[tex]y=-\frac{2}{3}x[/tex] --------(1)
Slope of the line is,
[tex]m_1=-\frac{2}{3}[/tex]
By the property of perpendicular lines,
If the two lines having slopes [tex]m_1[/tex] and [tex]m_2[/tex] are perpendicular to each other, [tex]m_1\times m_2=-1[/tex]
Therefore, slope of the line perpendicular to the line (1) will be,
[tex]-\frac{2}{3}\times m_2=-1[/tex]
[tex]m_2=\frac{3}{2}[/tex]
If this line passes through a point (4, -8) equation of the line will be,
[tex]y-y'=m_2(x-x')[/tex]
[tex]y+8=\frac{3}{2}(x-4)[/tex]
[tex]y=\frac{3}{2}x-6-8[/tex]
[tex]y=\frac{3}{2}x-14[/tex]