Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis.
A. \(\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}\) B. \(\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}\) C. \(\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}\) D. \(\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}\)
Correct Answer: A
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; \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\) \([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\) \([\frac{3}{8}, \frac{-3}{2}]\)