Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)
A. 7y + 4x + 5 = 0 B. 7y - 4x - 5 = 0 C. 3y - 5x + 2 = 0 D. 3y + 5x - 2 = 0
Correct Answer: A
Explanation
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. \(1 = \frac{-4}{7} (-3) + b\) \(1 = \frac{12}{7} + b\) \(b = \frac{-5}{7}\) \(\therefore y = \frac{-4}{7} x - \frac{5}{7}\) \(7y + 4x + 5 = 0\)