The midpoint of the segment of the line y = 4x + 3 which lies between the x-ax 1 is and the y-ax 1 is
(\(\frac{3}{2}\), \(\frac{3}{2}\))
(\(\frac{2}{3}\), \(\frac{3}{2}\))
(\(\frac{3}{8}\), \(\frac{3}{2}\))
(-\(\frac{3}{8}\), \(\frac{3}{2}\))
Correct answer is D
y = 4x + 3
when x = 0, y = 3 \(\to\) (0, 3)
when y = 0, x = -\(\frac{3}{4}\) \(\to\) (\(\frac{3}{4}\), 0)
mid-point \(\frac{0 + (-{\frac{3}{4}})}{2}\), \(\frac{3 + 0}{4}\)
-\(\frac{3}{8}\), \(\frac{3}{2}\)