Question

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

Answer 1

Explanation:

Answer 2

14.
`y = -2x -1`

15.
`y = -(3)/(4)x +3`

16.
`y= 4x + 9`

17.
`y = -(5)/(3)x -2`

18.
`y= -(2)/(3)x -5`

19.
`y = 4x -3`

20.
`y = -3x -7`

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) = -(3)/(4)`

Equation in
`slope- intercept` form is:

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

Substitute values for y1, m and x1

`y - 6 = -(3)/(4)(x - (-4))`

`y - 6 = -(3)/(4)(x +4)`

`y - 6 = -(3)/(4)x -3`

Collect Like Terms

`y = -(3)/(4)x -3 + 6`

`y = -(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 = (y_2 - y_1)/(x_2 - x_1)`

`m = (1 - (-11))/(-2 - (-5))`

`m = (1 +11)/(-2 +5)`

`m = (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 = (y_2 - y_1)/(x_2 - x_1)`

`m = (-7 - 8)/(3- (-6))`

`m = (-7 - 8)/(3+6)`

`m = (-15)/(9)`

`m = -(5)/(3)`

Equation in
`slope- intercept` form is:

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

Substitute values for y1, m and x1

`y - 8 = -(5)/(3)(x -(-6))`

`y - 8 = -(5)/(3)(x +6)`

`y - 8 = -(5)/(3)x -10`

Collect Like Terms

`y = -(5)/(3)x -10 + 8`

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

18.

Given

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

`y = -(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 = -(2)/(3)`

and

`m_1 = m_2 = -(2)/(3)`

Equation in
`slope\ intercept\ form` is:

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

Substitute values for y1, m and x1

`y - (-1) = -(2)/(3)(x - (-6))`

`y +1 = -(2)/(3)(x +6)`

`y +1 = -(2)/(3)x -4`

`y= -(2)/(3)x -4 - 1`

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

19.

Given

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

`y = -(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 = -(1)/(m_2)`

Where
`m_2` represents the slope

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

`m_2 = -(1)/(4)`

and

`m_1 = -(1)/(m_2)`

`m_1 = -1/(-1)/(4)`

`m_1 = -1*(-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 = (y_2 - y_1)/(x_2 - x_1)`

`m = (7 - 3)/(2 - (-10))`

`m = (7 - 3)/(2 +10)`

`m = (4)/(12)`

`m = (1)/(3)`

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

`m_1 = -(1)/(m_2)`

`m_1 = -1/(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 = [(1)/(2)(x_1+x_2),(1)/(2)(y_1+y_2)]`

`Midpoint = [(1)/(2)(-10+2),(1)/(2)(3+7)]`

`Midpoint = [(1)/(2)(-8),(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`