Question

What is the midpoint of a line segment with endpoints at (3,-1) and (8, -4)?

Answer 1

Answer:

Step-by-step explanation:($(\frac{3+8}{2} ,\frac{-1-4}{2} )\\\\(\frac{11}{2} ,\frac{-5}{2} )$

Answer 2

The midpoint of a line segment with endpoints at (3,-1) and (8, -4) is $\left(\frac{11}{2},-\frac{5}{2}\right)$

Solution:

We have been given 2 end points of a line which are: (3,-1) and (8, -4)

The midpoint of a line segment is half way from both the ends of the line segment.

The formula for midpoint for the two points $P(x_1, y_1) \text{and} Q(x_2, y_2)$ is given as:

$\text {Midpoint}=(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$

Here P(3, -1) and Q(8, -4)

$\text { So } x_{1}=3 ; x_{2}=8 ; y_{1}=-1 ; y_{2}=-4$

Plugging in values in above formula, we get

$\begin{array}{l}{\text {Midpoint}=(\frac{3+8}{2}, \frac{-1-4}{2}})\\\\ {\text {Midpoint}=(\frac{11}{2}, \frac{-5}{2}})\end{array}$

Hence, the midpoint of the line segment is $\left(\frac{11}{2},-\frac{5}{2}\right)$