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

Answer:

14. $y = -2x - 1$

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

16. $y = 4x + 9$

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

18. y = -⅔x - 5

19. y = 4x - 3

29. $y = -3x - 7$

Step-by-sep explanation:

✍️Equation of a line in slope-intercept form is given as $y = mx + b$. Where, m is the slope, and b is the y-intercept.

The following shows how to derive an equation of a line in slope-intercept form, if we are given a point and slope of the line between two points

14. (-7, 13); slope = -2.

Substitute x = -7, y = 13, and m = -2 into $y = mx + b$.

$13 = (-2)(-7) + b$

$13 = 14 + b$

Subtract 14 from both sides

$13 - 14 = b$

$-1 = b$

Substitute m = -2 and b = -1 in $y = mx + b$ to derive the equation:

✅$y = -2x + (-1)$

$y = -2x - 1$

15. (-4, 6); slope = -¾

Substitute x = -4, y = 6, and m = -¾ into $y = mx + b$.

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

$6 = 3 + b$

Subtract 3 from both sides

$6 - 3 = b$

$3 = b$

Substitute m = -¾ and b = 3 in $y = mx + b$ to derive the equation:

✅$y = -\frac{3}{4}x + 3$

✍️The following shows how to derive an equation of a line in slope-intercept form, if we are given two points on the line, only.

16. (-5, -11) and (-2, 1)

Find the slope

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 -(-11)}{-2 -(-5)} = \frac{12}{3} = 4$

Substitute x = -5, y = -11, and m = 4 into into $y = mx + b$.

$-11 = (4)(-5) + b$

$-11 = -20 + b$

Add 20 to both sides

$-11 + 20 = b$

$9 = b$

Substitute m = 4 and b = 9 in $y = mx + b$ to derive the equation:

✅$y = 4x + 9$

17. (-6, 8) and (3, -7)

Find the slope

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 8}{3 -(-6)} = \frac{-15}{9} = -\frac{5}{3}$

Substitute x = -6, y = 8, and m = -⁵/3 into into $y = mx + b$.

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

$8 = 10 + b$

Subtract 10 from both sides

$8 - 10 = b$

$-2 = b$

Substitute m = -⁵/3 and b = -2 in $y = mx + b$ to derive the equation:

✅$y = -\frac{5}{3}x - 2$

18. Given that the line that passes through the point, (-6, -1) is parallel to y = -⅔x + 1, therefore, it would have the same slope value as -⅔, as the line it is parallel to.

So, using a point (-6, -1) and slope (m) = -⅔, we can generate the equation of the line in slope-intercept form as follows:

Substitute x = -6, y = -1, and m = -⅔ in y = mx + b, to find b.

-1 = (-⅔)(-6) + b

-1 = 4 + b

-1 - 4 = b

-5 = b

Substitute m = -⅔ and b = -5 in y = mx + b, to generate the equation of the line.

✅y = -⅔x - 5

19. Given that the line that passes through the point, (-2, -11) is perpendicular to y = -¼x + 2, therefore, it would have the a slope value that is the negative reciprocal of the slope of the line that it is perpendicular to.

The slope of the line that it is perpendicular to is -¼. Therefore, the slope of the line that passes through (-2, -11), would be 4. (4 is the negative reciprocal of -¼)

So, using the point (-2, -11) and slope (m) = 4, we can generate the equation of the line in slope-intercept form as follows:

Substitute x = -2, y = -11, and m = 4 in y = mx + b, to find b.

-11 = (4)(-2) + b

-11 = -8 + b

-11 + 8 = b

-3 = b

Substitute m = 4 and b = -3 in y = mx + b, to generate the equation of the line.

✅y = 4x - 3

20. To solve this problem, first find the slope of the line that runs through A(-10, 3) and B(2, 7):

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 3}{2 -(-10)} = \frac{4}{12} = \frac{1}{3}$

Next, find the coordinates of the midpoint of AB.

$M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

$M(\frac{-10 + 2}{2}, \frac{3 + 7}{2})$

$M(\frac{-8}{2}, \frac{10}{2})$

$M(-4, 5)$

Since the slope of AB is ⅓, the slope of line l that is perpendicular to AB would be the negative reciprocal of ⅓.

Therefore, the slope of line l = -3.

Since line l, bisects AB, therefore, the coordinate of the mid-point of AB is also the same as a coordinate point on line l.

So therefore, using the midpoint, (-4, 5) and slope, m = -3, we can generate an equation for line l as follows:

Substitute x = -4, y = 5, and m = -3 into $y = mx + b$.

$5 = (-3)(-4) + b$

$5 = 12 + b$

Subtract 12 from both sides

$5 - 12 = b$

$-7 = b$

Substitute m = -3 and b = -7 in $y = mx + b$ to derive the equation:

✅$y = -3x + (-7)$

$y = -3x - 7$

Answer 2

Answer:

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

$y = -3x - 12 +5$

$y = -3x -7$