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JAMB Mathematics 2014 Paper

Question : 27

Total: 47

Solution:

y - 4x + 3 = 0

When y = 0, 0 - 4x + 3 = 0

Then -4x = -3

x = 3/4

So the line cuts the x-axis at point (3/4, 0).

When x = 0, y - 4(0) + 3 = 0

Then y + 3 = 0

y = -3

So the line cuts the y-axis at the point (0, -3)

Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

\left[ \frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2) \right]

\left[ \frac{1}{2}\left( \frac{3}{4} + 0 \right), \frac{1}{2}\left( 0 + -3 \right) \right]

\left[ \frac{1}{2}\left( \frac{3}{4} \right), \frac{1}{2}(-3) \right]

\left[ \frac{3}{8}, \frac{-3}{2} \right]

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