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Flauer [41]
3 years ago
5

Give the equation of the line perpendicular to the line through (3, 2) and (-2, 4) that

Mathematics
1 answer:
elena55 [62]3 years ago
4 0

Answer:

y=(5/2)x - 3/2

Step-by-step explanation:

We can start by finding the equation of the line given from the points. To do this, we will calculate the slope first then the y-intercept.

Line Equation: y=mx+b

Finding the slope value,

m= (y2-y1)/(x2-x1)

m=(4-2)/(-2-3)

m=-2/5

Finding the y-intercept,

y=(-2/5)x+b

Substituting one of the two coordinate pairs into the equation,

2=(-2/5)3+b

2=-6/5 +b

b=16/5

Thus the equation of the line is y=(-2/5)x + 16/5

To find the line perpendicular to this line, we have to take the negative reciprocal of the slope. So our equation for the perpendicular line is,

y=(5/2)x + b

Since the perpendicular line passes through (1,1), we can substitute this point to get the value of b,

1=(5/2)1 +b

1=5/2 + b

b= -3/2

Therefore, the equation of the line is y=(5/2)x - 3/2

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Directions: Calculate the area of a circle using 3.14x the radius
Leokris [45]

\qquad\qquad\huge\underline{{\sf Answer}}♨

As we know ~

Area of the circle is :

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 4.4\div 2

\qquad \sf  \dashrightarrow \:r = 2.2 \: mm

Now find the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.2)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {4.84}^{}

\qquad \sf  \dashrightarrow \:area  \approx 15.2 \:  \: mm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 3.7 \div 2

\qquad \sf  \dashrightarrow \:r = 1.85 \:  \: cm

Bow, calculate the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{}

\qquad \sf  \dashrightarrow \:area  \approx 10.75 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 3 </h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 68.89

\qquad \sf  \dashrightarrow \:area \approx216.31 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 4</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 5.8 \div 2

\qquad \sf  \dashrightarrow \:r = 2.9 \:  \: yd

now, let's calculate area ~

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.9)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  8.41

\qquad \sf  \dashrightarrow \:area  \approx26.41 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

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Now, let's calculate area ~

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2}

\qquad \sf  \dashrightarrow \:3.14  \times 0.25

\qquad \sf  \dashrightarrow \:area \approx0.785 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 6</h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(8)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 64

\qquad \sf  \dashrightarrow \:area = 200.96 \:  \: yd {}^{2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

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3 years ago
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Step-by-step explanation:

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