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3 years ago

What are the coordinates of the midpoint of the segment whose endpoints are A(−1, 9) and B(3, 8) ? Answer in decimals

Mathematics

2 answers:

maxonik [38]3 years ago

5 0

Midpoint formula: (x1+x2/2 , y1+y2/2)
So, (-1+3/2 , 9+8/2)
-2/2 = -1 17/2 = 8.5
Answer: (-1,8.5)

Lorico [155]3 years ago

5 0

Answer:

Mid point of line segment whose endpoints are A(−1, 9) and B(3, 8) is (1,8.5)

Step-by-step explanation:

We are given endpoints of a line segment as A(−1, 9) and B(3, 8) and we have to find the mid points of the given line segment

Mid point can be calculated using mid point formula.

Mid point formula stated as for a given line segment having end points A( $x_1,y_1$ ) and B( $x_2,y_2$ ) , we can calculate coordinates of mid point using,

$(x_3,y_3)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Where $(x_3,y_3)$ is the mid point.

We are given $x_1=-1$ , $y_1=9$ , $x_2=3$ , $y_2=8$

Put the values above, we get,

$(x_3,y_3)=(\frac{-1+3}{2},\frac{9+8}{2})$

$\Rightarrow (x_3,y_3)=(\frac{2}{2},\frac{17}{2})$

$\Rightarrow (x_3,y_3)=(1,8.5)$

Thus, mid point of line segment whose endpoints are A(−1, 9) and B(3, 8) is (1,8.5).

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8 0

3 years ago

Read 2 more answers

1. Two object accumulated a charge of

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Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$