Question

Find the coordinated points which quadrasects the line segments joining the points A(-3,6) and B (4,-2)

Answer 1

Answer:

The coordinated points which quadrasects the line segments joining the points A and B are $\left(-\frac{5}{4},4\right)$, $\left(\frac{1}{2}, 2 \right)$ and $\left(\frac{9}{4},0 \right)$.

Step-by-step explanation:

Let $A(x,y) =(-3,6)$ and $B(x,y) = (4,-2)$, when the line segment is quadrasected, it means that segment is divided into four equal parts. The locations are determined by the following expressions:

$\vec R_{1} = \vec A + \frac{1}{4}\cdot \overrightarrow{AB}$ (1)

$\vec R_{2} = \vec A + \frac{1}{2}\cdot \overrightarrow{AB}$ (2)

$\vec R_{3} = \vec A + \frac{3}{4}\cdot \overrightarrow{AB}$ (3)

Where:

$\overrightarrow{AB} = B(x,y)-A(x,y)$

$\overrightarrow{AB} = (4,-2) - (-3,6)$

$\overrightarrow{AB} = (7,-8)$

The coordinated points which quadrasects the line segments joining the points A and B are, respectively:

$\vec R_{1} = (-3,6)+\frac{1}{4} \cdot (7,-8)$

$\vec R_{1} = \left(-\frac{5}{4},4\right)$

$\vec R_{2} = (-3,6)+\frac{1}{2} \cdot (7,-8)$

$\vec R_{2} = \left(\frac{1}{2},2 \right)$

$\vec R_{3} = (-3,6)+\frac{3}{4} \cdot (7,-8)$

$\vec R_{3} = \left(\frac{9}{4},0\right)$

The coordinated points which quadrasects the line segments joining the points A and B are $\left(-\frac{5}{4},4\right)$, $\left(\frac{1}{2}, 2 \right)$ and $\left(\frac{9}{4},0 \right)$.