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WAEC Mathematics 2019 Paper

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Question : 34

Total: 50

Find the equation of a straight line passing through the point (1, -5) and having gradient of

\;\frac{3}{4}

3x + 4y - 23 = 0
3x + 4y +23 = 0
3x - 4y +23 = 0
3x - 4y - 23 = 0

Solution:

$\;\frac{3}{4} = \;\frac{y+5}{x-1}$

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

3x - 3 = 4y + 20

3x - 4y - 23 = 0

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