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Find the equation of a line perpendicular to the line 4y = 7...

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 is A

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.

\(1 = \frac{-4}{7} (-3) + b\)

\(1 = \frac{12}{7} + b\)

\(b = \frac{-5}{7}\)

\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)

\(7y + 4x + 5 = 0\)