How do you find the midpoint of (-4, 9) and (1, -3)?

Alberto Calhoun
2022-11-20
Answered

How do you find the midpoint of (-4, 9) and (1, -3)?

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cismadmec

Answered 2022-11-21
Author has **22** answers

Step 1

The points are:

$(-4,9)={{x}_{1},{y}_{1}}$

$(1,-3)={{x}_{2},{y}_{2}}$

The formula to calculate the midpoint is:

$(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})$

$=(\frac{-4+1}{2},\frac{9+(-3)}{2})$

$=(\frac{-3}{2},\frac{6}{2})$

$=(\frac{-3}{2},3)$

The points are:

$(-4,9)={{x}_{1},{y}_{1}}$

$(1,-3)={{x}_{2},{y}_{2}}$

The formula to calculate the midpoint is:

$(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})$

$=(\frac{-4+1}{2},\frac{9+(-3)}{2})$

$=(\frac{-3}{2},\frac{6}{2})$

$=(\frac{-3}{2},3)$

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What is the midpoint of the line segment that joins points (4, -2) and (-2, 5)?

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Consider the hyperbolic 3-space $({\mathbb{H}}^{3},d{s}^{2})$ with

${\mathbb{H}}^{3}:=\{(x,y,z)\in {\mathbb{R}}^{3}|z>0\},\phantom{\rule{1em}{0ex}}d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}+d{z}^{2}}{{z}^{2}}$

Geodesics for this space are circular arcs normal to $\{z=0\}$ and vertical rays normal to $\{z=0\}$ .

${\mathbb{H}}^{3}:=\{(x,y,z)\in {\mathbb{R}}^{3}|z>0\},\phantom{\rule{1em}{0ex}}d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}+d{z}^{2}}{{z}^{2}}$

Geodesics for this space are circular arcs normal to $\{z=0\}$ and vertical rays normal to $\{z=0\}$ .

asked 2022-05-07

I wanted to derive the formula for the error in the basic midpoint rule.

$E(f)={\int}_{a}^{b}f[{\textstyle \frac{a+b}{2}},x](x-{\textstyle \frac{a+b}{2}})\phantom{\rule{thinmathspace}{0ex}}dx.$

but I don't see how:

$E(f)={\int}_{a}^{b}f[{\textstyle \frac{a+b}{2}},x](x-{\textstyle \frac{a+b}{2}})dx=(\ast )$ ?

$E(f)={\int}_{a}^{b}f[{\textstyle \frac{a+b}{2}},x](x-{\textstyle \frac{a+b}{2}})\phantom{\rule{thinmathspace}{0ex}}dx.$

but I don't see how:

$E(f)={\int}_{a}^{b}f[{\textstyle \frac{a+b}{2}},x](x-{\textstyle \frac{a+b}{2}})dx=(\ast )$ ?

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Lines AC and CE are perpendicular. D is the midpoint of line segment EC and B is the midpoint of line segmentAC. Segments AD and BE are drawn and intersect at point F. CD = 15 AB = 15. Determine the area ofthe triangular region DEF.

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What is the distance of (1,4) and (5,2)?

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Point A is located at $(2,-6)$ , and D is located at (-6, 8). What are the coordinates of the point that lies halfway between A and D?

asked 2022-07-04

I've gotten to the point where I have the following equation:

$2{x}_{2}{x}_{m}-2{x}_{m}{x}_{1}+2{y}_{2}{y}_{m}-2{y}_{m}{y}_{1}={x}_{2}^{2}-{x}_{1}^{2}+{y}_{2}^{2}-{y}_{1}^{2}$

Would it be mathematically correct to split this into the following two equations

$\{\begin{array}{l}2{x}_{2}{x}_{m}-2{x}_{m}{x}_{1}={x}_{2}^{2}-{x}_{1}^{2}\\ 2{y}_{2}{y}_{2}-2{y}_{m}{y}_{1}={y}_{2}^{2}-{y}_{1}^{2}\end{array}$

and treat them as a system of equations? If so, then how would I go about doing this?

$2{x}_{2}{x}_{m}-2{x}_{m}{x}_{1}+2{y}_{2}{y}_{m}-2{y}_{m}{y}_{1}={x}_{2}^{2}-{x}_{1}^{2}+{y}_{2}^{2}-{y}_{1}^{2}$

Would it be mathematically correct to split this into the following two equations

$\{\begin{array}{l}2{x}_{2}{x}_{m}-2{x}_{m}{x}_{1}={x}_{2}^{2}-{x}_{1}^{2}\\ 2{y}_{2}{y}_{2}-2{y}_{m}{y}_{1}={y}_{2}^{2}-{y}_{1}^{2}\end{array}$

and treat them as a system of equations? If so, then how would I go about doing this?