Mathematics, 13.11.2020 21:00 Derick21

Quiz 3-3 Parallel and Perpendicular Lines on the Coordinate Plane (Gina Wilson All Things Algebra 2014-2019) need these for a quiz please!

Quiz 3-3 Parallel and Perpendicular Lines on the Coordinate Plane (Gina Wilson All Things Algebra 2

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Answer from: nyandog50

14. y = -2x -1

15. $y = -\frac{3}{4}x +3$

16. y= 4x + 9

17. $y = -\frac{5}{3}x -2$

18. $y= -\frac{2}{3}x -5$

19. y = 4x -3

20. y = -3x -7

Step-by-step explanation:

Solving (14):

Given

(x_1,y_1) = (-7,13)

Slope (m) = -2

Equation in slope- intercept form is:

y - y_1 = m(x-x_1)

Substitute values for y1, m and x1

y - 13 = -2(x -(-7))

y - 13 = -2(x +7)

y - 13 = -2x -14

Collect Like Terms

y = -2x -14 + 13

y = -2x -1

Solving (15):

Given

(x_1,y_1) = (-4,6)

$Slope (m) = -\frac{3}{4}$

Equation in slope- intercept form is:

y - y_1 = m(x-x_1)

Substitute values for y1, m and x1

$y - 6 = -\frac{3}{4}(x - (-4))$

$y - 6 = -\frac{3}{4}(x +4)$

$y - 6 = -\frac{3}{4}x -3$

Collect Like Terms

$y = -\frac{3}{4}x -3 + 6$

$y = -\frac{3}{4}x +3$

Solving (16):

Given

(x_1,y_1) = (-5,-11)

(x_2,y_2) = (-2,1)

First, we need to calculate the $slope\ (m)$

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{1 - (-11)}{-2 - (-5)}$

$m = \frac{1 +11}{-2 +5}$

$m = \frac{12}{3}$

m = 4

Equation in slope- intercept form is:

y - y_1 = m(x-x_1)

Substitute values for y1, m and x1

y - (-11) = 4(x -(-5))

y +11 = 4(x+5)

y +11 = 4x+20

Collect Like Terms

y= 4x + 20 - 11

y= 4x + 9

Solving (17):

Given

(x_1,y_1) = (-6,8)

(x_2,y_2) = (3,-7)

First, we need to calculate the $slope\ (m)$

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{-7 - 8}{3- (-6)}$

$m = \frac{-7 - 8}{3+6}$

$m = \frac{-15}{9}$

$m = -\frac{5}{3}$

Equation in slope- intercept form is:

y - y_1 = m(x-x_1)

Substitute values for y1, m and x1

$y - 8 = -\frac{5}{3}(x -(-6))$

$y - 8 = -\frac{5}{3}(x +6)$

$y - 8 = -\frac{5}{3}x -10$

Collect Like Terms

$y = -\frac{5}{3}x -10 + 8$

$y = -\frac{5}{3}x -2$

18.

Given

(x_1,y_1) = (-6,-1)

$y = -\frac{2}{3}x+1$

Since the given point is parallel to the line equation, then the slope of the point is calculated as:

m_1 = m_2

Where m_2 represents the slope

Going by the format of an equation, y = mx + b ; by comparison

$m = -\frac{2}{3}$

and

$m_1 = m_2 = -\frac{2}{3}$

Equation in $slope\ intercept\ form$ is:

y - y_1 = m(x-x_1)

Substitute values for y1, m and x1

$y - (-1) = -\frac{2}{3}(x - (-6))$

$y +1 = -\frac{2}{3}(x +6)$

$y +1 = -\frac{2}{3}x -4$

$y= -\frac{2}{3}x -4 - 1$

$y= -\frac{2}{3}x -5$

19.

Given

(x_1,y_1) = (-2,-11)

$y = -\frac{1}{4}x+2$

Since the given point is parallel to the line equation, then the slope of the point is calculated as:

$m_1 = -\frac{1}{m_2}$

Where m_2 represents the slope

Going by the format of an equation, y = mx + b ; by comparison

$m_2 = -\frac{1}{4}$

and

$m_1 = -\frac{1}{m_2}$

$m_1 = -1/\frac{-1}{4}$

$m_1 = -1*\frac{-4}{1}$

m_1 = 4

Equation in $slope\ intercept\ form$ is:

y - y_1 = m(x-x_1)

(x_1,y_1) = (-2,-11)

Substitute values for y1, m and x1

y - (-11) = 4(x - (-2))

y +11 = 4(x +2)

y +11 = 4x +8

Collect Like Terms

y = 4x + 8 - 11

y = 4x -3

20.

Given

(x_1,y_1) = (-10,3)

(x_2,y_2) = (2,7)

First, we need to calculate the slope of the given points

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{7 - 3}{2 - (-10)}$

$m = \frac{7 - 3}{2 +10}$

$m = \frac{4}{12}$

$m = \frac{1}{3}$

Next, we determine the slope of the perpendicular bisector using:

$m_1 = -\frac{1}{m_2}$

$m_1 = -1/\frac{1}{3}$

m_1 = -3

Next, is to determine the coordinates of the bisector.

To bisect means to divide into equal parts.

So the coordinates of the bisector is the midpoint of the given points;

$Midpoint = [\frac{1}{2}(x_1+x_2),\frac{1}{2}(y_1+y_2)]$

$Midpoint = [\frac{1}{2}(-10+2),\frac{1}{2}(3+7)]$

$Midpoint = [\frac{1}{2}(-8),\frac{1}{2}(10)]$

Midpoint = (-4,5)

So, the coordinates of the midpoint is:

(x_1,y_1) = (-4,5)

Equation in slope- intercept form is:

y - y_1 = m(x-x_1)

Substitute values for y1, m and x1: m_1 = -3 & (x_1,y_1) = (-4,5)

y - 5 = -3(x - (-4))

y - 5 = -3(x +4)

y - 5 = -3x-12

Collect Like Terms

y = -3x - 12 +5

y = -3x -7

Answer from: Quest

5would be the answer- explanation

Answer from: Quest

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 = (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4 we need to find some relationship between a and b.

let's try ab = bc

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2

br = sqrt(3) * a

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram.

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a

cp = sqrt(3)*a

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt 3) * a which agrees with the other two.

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Quiz 3-3 Parallel and Perpendicular Lines on the Coordinate Plane (Gina Wilson All Things Algebra 20...