Please answer the following algebra I math problems showing ALL steps?

Question

Wrte in slope intercept form the equation of the line through the given point and perpendicular to the given line:

a. (-1,-8), y= -3/4x + 4
b. (6, -8), y= -6x - 9

Write the equation of the line in standard form. Use interger coefficients.

y=5/6x + 5/12

Thanks

Answer 1

a) (-1,-8), y=-3/4x + 4
for line perpendicular to y=-3/4x + 4
m*(-3/4) = -1
m = 4/3
the line is y = 4/3x + c
for (-1,-8), -8 = 4/3(-1)+c
c = -20/3

So, the line y=4/3x - 20/3

b) (6,-8), y=-6x - 9
for line perpendicular to y=-6x - 9
m*(-6) = -1
m = 1/6
the line is y = 1/6x + c
for (6,-8), -8 = 1/6(6)+c
c = -9

So, the line y=1/6x - 9

Answer 2

a) slope of the line = 4/3 ( as the line is perpendicular so the product of the slopes of the two lines is -1)
-8 = 4/3 * -1 + c
-8 = -4/3 +c
-24 = -4 + 3c
-20 = 3c
c = -20/3
equation : y = 4x/3 - 20/3
3y = 4x - 20
4x - 3y = 20

b) slope of the line = 1/6
-8 = 6 * 1/6 +c
c = -9
equation : y= 1x/6 -9
6y = x - 54
x - 6y = 54

Answer 3

if the graph is y=ax+b, then the slope of the perpendicular is -1/a!

the reason is as following
if the angle between the graph and X-coordinate is A!
then the perpendicular's angle is 90+A
so the slope of the perpendicular is
tan(90+A)=-cotA=-1/tanA=-1/a

a) the slope the perpendicular is -1/(-3/4)=4/3
y1=4/3x+b!
(-1,-8)->y1, then
-8=-4/3+b
b=-20/3
so
y1=4/3x-20/3

b) the slope of the perpendicular is -1/(-6)=1/6
so y1=1/6x+b
(6,-8)->y1 then,
-8=6*1/6+b
-8=1+b
b=-9
so.
y1=1/6x-9

Answer 4

slope intercept form of a line is y=mx+c

a.slope of a given line y=mx+c is "m". so slope of the given line is -3/4
to find perpendicular line slope:m1*m2= -1
so slope of required line is 4/3
so eq. of a line is given by : y-y1=m(x-x1)

y+8=4/3(x+1)
3y+24=4x+4 is the answer {y1&x1 are the given pts.}

proceed the other in the same way!!!

std form {i'am not sure} but i thk it is 12y=10x+5{jst solve by taking lcm}

Answer 5

normal= -1/m

where m= y1-y/x1-x

so these two lines given are in the form

y=mx+b where a is the slope

so for the lines y= -3/4x+4

m=-3/4

normal line= -1/(-3/4)= -1*4/-3

= 4/3

the equation will be such in correspondance to given coordinates (-1,-8)

a. y= 4/3(x+1) -8
------------------------------

b. m= -6

normal= -1/-6= 1/6

so the equation of such a line in throught coordinates (6,-8)

y= 1/6(x-6) -8

----------------------------------------------

6y=5x +5/2

12y=10x+5

12y-10x=5(this is in integer format but I am not really sure if it is in standard form, please rearrange if I made a mistake in my manner of standard form)

Answer 6

a. pt= (-1,-8) m= 4/3
ans: y = 4/3x -20/3

b. pt =( 6.-8) m= 1/6
ans: y= 1/6x -9

Answer 7

do your own homework.