JAMB - Mathematics (2014 - No. 27)
Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis.
\(\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}\)
\(\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}\)
\(\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}\)
Explanation
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;
\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)
\([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\)
\([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\)
\([\frac{3}{8}, \frac{-3}{2}]\)
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;
\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)
\([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\)
\([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\)
\([\frac{3}{8}, \frac{-3}{2}]\)
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