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

Assessment: JAMB UTME - Mathematics - 2019 Subject: General Mathematics

Question 1 Report

7y + 4x + 5 = 0

7y - 4x - 5 = 0

3y - 5x + 2 = 0

3y + 5x - 2 = 0

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)

∴

Using the equation of line

y = m x + 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}

∴ y = \frac{- 4}{7} x - \frac{5}{7}

7 y + 4 x + 5 = 0

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