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2 years ago

Write an equation in point-slope form of the line that passes through (-4,1) and (4,3).

Mathematics

1 answer:

Step2247 [10]2 years ago

5 0

Answer:

An equation in point-slope form of the line that passes through (-4,1) and (4,3) will be:

$y-1=\frac{1}{4}\left(x+4\right)$

Step-by-step explanation:

Given the points

(-4,1)

(4,3)

Finding the slope between the points (-4,1) and (4,3)

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-4,\:1\right),\:\left(x_2,\:y_2\right)=\left(4,\:3\right)$

$m=\frac{3-1}{4-\left(-4\right)}$

Refine

$m=\frac{1}{4}$

Point slope form:

$y-y_1=m\left(x-x_1\right)$

where

m is the slope of the line

(x₁, y₁) is the point

in our case,

m = 1/4

(x₁, y₁) = (-4,1)

substituting the values m = 1/4 and the point (-4,1) in the point slope form of line equation.

$y-y_1=m\left(x-x_1\right)$

$y-1=\frac{1}{4}\left(x-\left(-4\right)\right)$

$y-1=\frac{1}{4}\left(x+4\right)$

Thus, an equation in point-slope form of the line that passes through (-4,1) and (4,3) will be:

$y-1=\frac{1}{4}\left(x+4\right)$

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2 years ago

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2 years ago

Assume z = x + iy, then find a complex number z satisfying the given equation. d. 2z8 – 2z4 + 1 = 0

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Answer: complex equations has n number of solutions, been n the equation degree. In this case:

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$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}$

Step-by-step explanation:

I start with a variable substitution:

$Z^{4} = X$

Then:

$2X^{2}-2X+1=0$

Solving the quadratic equation:

$X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}$

$X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.$

Replacing for the original variable:

$Z=\sqrt[4]{0,5+0,5i}$

or $Z=\sqrt[4]{0,5-0,5i}$

Remembering that complex numbers can be written as:

$Z=a+ib=|Z|e^{ic}$

Using this:

$Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.$

Solving for the modulus and the angle:

$Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.$

The possible angle respond to:

$RAng_{12...n} =\frac{Ang +360*(i-1)}{n}$

Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"

In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º

Obtaining 8 different angles, therefore 8 different solutions.

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2 years ago