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Gwar [14]
3 years ago
7

What is the equation of the line that passes through the point (1,3) and has a slope of -- 32

Mathematics
1 answer:
Korolek [52]3 years ago
5 0

Answer:

The equation of the line that passes through the point (1,3) and has a slope  of -32.

y=-32x+35

Please check the attached graph too.

Step-by-step explanation:

Given

  • Point (1, 3)
  • The slope m = -32

The 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 = -32

(x₁, y₁) = (1, 3)

substituting m = -32 and (x₁, y₁) = (1, 3) in the point-slope form of the line equation

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

y-3=-32\left(x-1\right)

Add 3 to both sides

y-3+3=-32\left(x-1\right)+3

Simplify

y=-32x+35

Therefore, the equation of the line that passes through the point (1,3) and has a slope  of -32.

y=-32x+35

Please check the attached graph too.

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Answer:

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Step-by-step explanation:

Given data

We are told that

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What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?
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The side lengths of triangle are 6 units, 8 units and 10 units.

<u>SOLUTION: </u>

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)  

We know that, distance between two points P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right) is given by  

P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Now,  

\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}

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Hello! 

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