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soldi70 [24.7K]
2 years ago
5

What is the midpoint of the segment shown below?

Mathematics
2 answers:
deff fn [24]2 years ago
8 0

The midpoint formula is basically (averaging the x coordinates, averaging the y coordinates).

Point A: (3, 7)

Point B: (2, -1)

Midpoint x: (3 + 2) / 2 = 5 / 2

Mindpoint y: (7 - 1) / 2 = 3

Therefore, the midpoint of the segment is choice C (5/2, 3)

elena-14-01-66 [18.8K]2 years ago
6 0

Answer:

Correct answer is Option C: (\frac{5}{2} , 3 )

Step-by-step explanation:

Given points (3, 7) and (2, -1), we can use the following Midpoint Formula to find out what is the halfway mark of the given segment:

M = (\frac{x1  +  x2}{2} , \frac{y1  + y2}{2} )

Let x1 = 3

x2 = 2

y1 = 7

y2 = -1

Plug in these values into the Midpoint formula:

M = (\frac{x1  +  x2}{2} , \frac{y1  + y2}{2} )

M = (\frac{3  +  2}{2} , \frac{-1 + 7}{2} )

M = (\frac{5}{2} , \frac{6}{2} ) =  (\frac{5}{2} , 3 )

Therefore, the midpoint of the given segment is (\frac{5}{2} , 3 ).

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Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is y=\frac{1}{3} x+2

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

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On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3  

And as the line passing through (-3,1) and is  perpendicular to 3x + y = -8, product of slopes of two line will be -1  as lies are perpendicular.

Let required slope = x  

\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}

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Equation of line passing through (x_1 , y_1) and having lope of m is given by

\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)

\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}

Substituting the values we get,

\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}

Hence the required equation of line is found using slope intercept form

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