Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

Answer Details

Equation: 4y = 7x + 3
$⟹y=\frac{7}{4}x+\frac{3}{4}$
Slope = coefficient of x = $\frac{7}{4}$
Slope of perpendicular line = $\frac{-1}{\frac{7}{4}}$
= $\frac{-4}{7}$
The perpendicular line passes (-3, 1)
$\therefore$ Using the equation of line $y=mx+b$
m = slope and b = intercept.
$y=\frac{-4}{7}x+b$
To find the intercept, substitute y = 1 and x = -3 in the equation.
$1=\frac{-4}{7}(-3)+b$
$1=\frac{12}{7}+b$
$b=\frac{-5}{7}$
$\therefore y=\frac{-4}{7}x-\frac{5}{7}$
$7y+4x+5=0$