Jambmaths question:
Calculate the midpoint of the line segment $y-4x+3=0$ which lies between the x –axis and y–axis
Option A:
$(-\tfrac{3}{2},\tfrac{2}{3})$
Option B:
$(-\tfrac{2}{3},\tfrac{3}{2})$
Option C:
$(\tfrac{3}{8},-\tfrac{3}{2})$
Option D:
$(\tfrac{3}{8},\tfrac{3}{2})$
Jamb Maths Solution:
$\begin{align} & y-4x+3=0 \\ & \text{When }x=0,\text{ }y=-3 \\ & ({{x}_{1}},{{y}_{1}})=(0,-3) \\ & \text{When }y=0,\text{ }x=\tfrac{3}{4} \\ & ({{x}_{2}},{{y}_{2}})=(\tfrac{3}{4},0) \\ & \text{Mid}-\text{point of a line segment} \\ & (x,y)=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\ & (x,y)=\left( \frac{0+\tfrac{3}{4}}{2},\frac{0-3}{2} \right) \\ & (x,y)=\left( \frac{3}{8},-\frac{3}{2} \right) \\\end{align}$
Jamb Maths Topic:
Year of Exam: