Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)
Equation: 4y = 7x + 3
\(\implies y = \frac{7}{4} x + \frac{3}{4}\)
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.
\(1 = \frac{-4}{7} (-3) + b\)
\(1 = \frac{12}{7} + b\)
\(b = \frac{-5}{7}\)
\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)
\(7y + 4x + 5 = 0\)
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