Find the equation of a straight line passing through the point (1, -5) and having gradient of \(\frac{3}{4}\)

  • A 3x + 4y - 23 = 0
  • B 3x + 4y +23 = 0
  • C 3x - 4y +23 = 0
  • D 3x - 4y - 23 = 0
waec 2019mathematics

The correct answer is D. 3x - 4y - 23 = 0

\(\frac{3}{4} = \frac{y + 5}{x - 1}\)

3(x - 1) = 4(y + 5)

3x - 3 = 4y + 20

3x - 4y - 23 = 0

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