algbera 2 homework problem! equation of the line

Question

1) find the equation of a line containing points (-2,8) and (3,5)

2) write an equation of the line parallel  to y=3x-2 through (1,-3)

3) write an equation of the line perpendicular  to x+y=-3 through (0,4)

Answer 1

1.

Given points are (-2,8) and (3,5)

Let A = (-2,8) = (x1,y1) and B = (3,5) = (x2,y2)

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

               = (5-8)/(3-(-2))

               = -3/3+2

          m = -3/5

The equation of line having slope m and point (x1,y1) is y-y1 = m(x-x1)

                                                                                 y-(-2) = -3/5(x-8)

                                                                                 5(y+2) = -3(x-8)

                                                                                 5y+10 = -3x+8

                                                                                 5y+3x = 8-10

                                                                                 5y+3x = -2

The equation of a line containing points (-2,8) and (3,5) is 5y+3x+2 = 0

Answer 2

2)

Given equation  y= 3x - 2 and point (1,-3)

so, slope(m) = 3

Write an equation of the line that passes through the given point and is parallel to the given line,
(1,-3), y = 3x - 2.

(1,-3) and slope(m) = 3

Write point - slope equation of the line is y - y1 = m(x - x1)

y - (-3) = 3(x -1)

Product of two same signs is positive

y + 3 = 3(x -1)

Apply distributive property:- a(b+c) = ab + ac

y + 3 = 3*x -3*1

y + 3 = 3x -3

Subtract 3 from each side

y + 3 - 3 = 3x - 3 - 3

y  = 3x - 6

The line equation is y  = 3x - 6.

Answer 3

3) The equation of line perpendicular to x + y= - 3

The slope of given line, m = -1/1=-1

The slope of the perpendicular line = -1.

(x1,y1) = (0,4).

The equation of perpendicular line is,

y-y1=m(x-x1)

y-4 = -1(x+0)

y-4 = -x

x + y - 4 =0

So the required equation of the perpendicular line is, x+y - 4=0

Answer 4

 

Given points are (-2,8) and (3,5)

Let A = (-2,8) = (x1,y1) and B = (3,5) = (x2,y2)

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

               = (5-8)/(3-(-2))

               = -3/3+2

          m = -3/5

The equation of line having slope m and point (x1,y1) is y-y1 = m(x-x1)

               ( x1,y1)  =   (-2,8)                                              y-(8) = -3/5(x-(-)2)

                    m = -3/5                                                          5(y-8) = -3(x+2)

                                                                                            5y-40 = -3x-6

                                                                                             5y+3x = -6+40

                                                                                              5y+3x = 34

                                                                                              3x+5y = 34

The equation of a line containing points (-2,8) and (3,5) is 3x+5y-34 = 0

Answer 5

3) The equation of line perpendicular to x + y= - 3

x+y = -3

add each side by -x

y = -x-3

The slope of given line by useing intercept form m = -1

The slope of the perpendicular line M = -1/m =-1/-1 = 1

(x1,y1) = (0,4).

The equation of perpendicular line is,

y-y1=1/M(x-x1)

y-4 = 1/1(x-0)

y-4 = x

x - y + 4 =0

So the required equation of the perpendicular line is, x-y + 4=0

 

Answer 6

3) The equation of line perpendicular to x + y= - 3

The slope of given line, m = -1/1=-1

The slope of the perpendicular line = -1/m = -1/-1 = 1

(x1,y1) = (0,4).

The equation of perpendicular line is,

y-y1=m(x-x1)

y-4 = 1(x+0)

y-4 = x

x - y + 4 =0

So the required equation of the perpendicular line is, x - y + 4 =0

 

Answer 7

 

 

1.

Given points are (-2,8) and (3,5)

Let A = (-2,8) = (x1,y1) and B = (3,5) = (x2,y2)

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

               = (5-8)/(3-(-2))

               = -3/3+2

          m = -3/5

The equation of line having slope m and point (x1,y1) is y-y1 = m(x-x1)

                                                                                 y-(8) = -3/5(x-(-2))

                                                                                 5(y-8) = -3(x+2)

                                                                                 5y-40 = -3x-6

                                                                                 5y+3x = -6+40

                                                                                 5y+3x = 34

The equation of a line containing points (-2,8) and (3,5) is    5y+3x = 34

 

 

Answer 8

 

3) The equation of line perpendicular to x + y= - 3

The slope of given line, m = -1/1=-1

The slope of the perpendicular line = -1/m

i.e, perpendicular slope is -1/-1 = 1

(x1,y1) = (0,4) ,m = 1

The equation of perpendicular line is,

y-y1=m(x-x1)

y-4 = 1(x+0)

y-4 = x

x - y + 4 =0

So the required equation of the perpendicular line is, x - y + 4=0

Answer 9

 

1.

Given points are (-2,8) and (3,5)

Let A = (-2,8) = (x1,y1) and B = (3,5) = (x2,y2)

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

               = (5-8)/(3-(-2))

               = -3/3+2

          m = -3/5

The equation of line having slope m and point (x1,y1) is y-y1 = m(x-x1)

                                                                                 y-(8) = -3/5(x-(-2))

                                                                                 5(y - 8) = -3(x+2)

                                                                                 5y - 40 = -3x - 6

                                                                                 5y+3x = - 6 + 40

                                                                                 5y+3x = 34

The equation of a line containing points (-2,8) and (3,5) is 5y+3x-34 = 0

 

Answer 10

1.

Given points are (-2,8) and (3,5)

Let A = (-2,8) = (x1,y1) and B = (3,5) = (x2,y2)

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

               = (5-8)/(3-(-2))

               = -3/3+2

          m = -3/5

The equation of line having slope m and point (x1,y1) is y-y1 = m(x-x1)

                                                                                 y-(8) = -3/5(x-(-2))

                                                                                 5(y - 8) = -3(x+2)

                                                                                 5y - 40 = -3x - 6

                                                                                 5y+3x = - 6 + 40

                                                                                 5y+3x = 34

The equation of a line containing points (-2,8) and (3,5) is 5y+3x-34 = 0

Answer 11

3)

The equation  of the line form = ax + by + c = 0

The equation of the perpendicular line form for above line = bx - ay + k = 0

Given equation x + y = -3 passing through (0 , 4)

a = 1, b = 1, c = 3 then

perpendicular line equation = 1x - 1y + k = 0 and

Passing through the point (0, 4) then

Substitute x = 0 , y = 4

The perpendicular line = 0 - 4 + k = 0

k = 4

There fore the perpendicular line = x - y + 4 = 0.