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3 years ago

8

The slope of the line below is -3 . Which of the following is the point slope form of the line ?

1 answer:

lidiya [134]3 years ago

4 0

Answer:

$y+2= -3(x-2)$

Step-by-step explanation:

The slope of the line below is $-3$

We need to find point slope form of the line.

point slope form of the line formula is

$y-y_1= m(x-x_1)$

Where m is the slope and (x1,y1) is the point on the line

From the given graph , point on the line is (2,-1)

slope is $-3$

Lets plug in the slope and the point in the formula

$y-(-2)= -3(x-2)$

$y+2= -3(x-2)$

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Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

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substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

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$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

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$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

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Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

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

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$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

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substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

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<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

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$b=5\ units$

$A=5^{2}=25\ units^{2}$

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

$P=4b$

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