How do you write the equations for the line that is perpendicular to the given line?

Question

and that passes through the given point?

(-5,5); y = -5x + 9

(-1,-4); y = 1/6x + 1

i think the formula is y - y1 = m (x - x1)
but i'm not sure.


can you please explain to me how you do it?

Answer 1

all you have to do is use the negative reciprocal of the slope in the new equation to form a perpindicular line. Here goes:

y = -5x + 9
slope= -5
slope of perpindicular line= 1/5
now apply that to a slope intercept equation using the coordinates of (-5,5) and solve for "b"
y = mx + b
5 = 1/5(-5) + b
25 = -1 + b
26 = b

FINAL ANSWER:
y = 1/5x + 26

now do the same thing for the second equation.
y = 1/6x + 1
slope =1/6
slope of perpindicular line= -6
now apply that to a slope intercept equation using the coordinates of (-1,-4) and solve for "b"
y = mx + b
-4 = -6(-1) + b
-4 = 6 + b
-10 = b

FINAL ANSWER:
y = -6x - 10

Answer 2

perpendicular lines have "negative reciprocal" slope.

all lines perpendicular to y = -5x + 9 have slope = 1/5, so using your point/slope form, the new line is
y-5 = (1/5)(x + 5)
y = x/5 + 6

perpendicular to (1/6)x + 1 thru (-1,-4) is
y + 4 = -6(x + 1)
y + 4 = -6x - 6
y = -6x - 10

Answer 3

Change the slope to the negative reciprocal, and then insert the given coordinates to find a particular solution and thus evaluate the constant. Here, y2 = 0.2 x2 + c [note that the +9 does not matter], and with 5 = 0.2(-5) + c, c = 6.

Answer 4

Well if this is homework I can't do it for you so all I can say is this
1. Take the Slope and write it's reciprocal
2. Make that negative.

That is the slope of a perpendicular line

Answer 5

per slpes are opp reciprocals
(-5,5); y = -5x + 9 has slope -5 so per slope =1/5
y=mx+b
5=(1/5)(-5)+b
b=6
y=(1/5)x+6

(-1,-4); y = 1/6x + 1 has slope 1/6, so perp slope =-6
-4=-6*-1+b
b=-10
y=-6x-10

Answer 6

1. find slope by (m1)(m2)= -1

2. use eq of line y-y1= m(x-x1)

3. put x1,y1 as given point & slope from 1