Find the equation of a line perpendicular to th... - JAMB Mathematics 2019 Question

the equation is 4y = 7x + 3

\(\implies y = \frac{7}{4} x + \frac{3}{4}\)

the slope = coefficient of x = \(\frac{7}{4}\)

the slope of a perpendicular line = \(\frac{-1}{\frac{7}{4}}\)

= \(\frac{-4}{7}\)

for perpendicular line passes (-3, 1)

\(\therefore\) using the equation of line \(y = mx + b\)

m = slope

b = intercept.

\(y = \frac{-4}{7} x + b\)

we substitute y = 1 and x = -3 in the equation to find the intercept:

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

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

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

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

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