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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:

Concept:

Use the point-gradient form of a straight line:

y - y_1 = m(x - x_1)

.

Explanation:

Given point

(1, -5)

and gradient

m = \frac{3}{4}

.

Substitute these values into the formula:

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

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

Multiply both sides by 4:

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

4y + 20 = 3x - 3

Rearrange to standard form:

3x - 4y - 23 = 0

This matches option D.

Answer:

3x - 4y - 23 = 0

(Option D)

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