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2 years ago

Write an equation of the line that passes through the given points (1 , 1.5) and (2, 3.5)

Mathematics

1 answer:

Nitella [24]2 years ago

5 0

Answer:

y=2x-0.5

Step-by-step explanation:

typically, the equation of the line is written in slope intercept form, which is y=mx+b, where m is the slope and b is the y intercept

first, we need to find the slope (m) of the line

the formula for slope calculated from two points is given as (y2-y1)/(x2-x1)

we have two points: (1,1.5) and (2,3.5) and we can label them

x1=1

y1=1.5

x2=2

y2=3.5

now plug into the formula:

m=(3.5-1.5)/(2-1)

m=2/1

m=2

the slope of the line is 2

here's our equation so far:

y=2x+b

now we need to find b

since the line will pass through the points (1, 1.5) and (2, 3.5), we can plug either one of them into the equation to solve for

Let's use (1, 1.5)

substitute 1.5 as y and 1 as x

1.5=2(1)+b

1.5=2+b

subtract 2 from both sides

-0.5=b

therefore, our equation is:

y=2x-0.5

hope this helps!

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Step-by-step explanation:

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Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

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

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

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

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

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$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

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GuDViN [60]

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Since you don't show your triangles, I'm guessing it is this one...?

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1. 90° counterclockwise rotation about the origin

2. Translation downward 2 units

Step-by-step explanation:

1. 90° counterclockwise rotation about the origin = (x, y) to (-y, x),

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