find an equation of the line through a which is perpendicular to ab?

Question

a(1,7) and b(5,1)

Answer 1

y-7 = [(5-1)/(7-1)](x-1) = (2/3)(x-1)

Reason: perpendicular slope = -1/(change of y / change of x)

Answer 2

First find the gradient of 'AB'.
The Gradient is :-
(1- 7)/ (5 - 1) =
-6/4 =
- 3/2
Gradients that are perpendicular to each other satisfy the equation : -
mm' = -1
Where 'm' and 'm'' are the two perpendicular gradients.#
With m = -3/2
Then -3/2 x m' = -1
m' = -1 x -2/3
m' = 2/3
So the line perp'r to AB has the form
y = 2x/3 + c
To find 'c' take the point A and substitute in its co-ordinates.
7 = 2(1)/3 + c
c = 7 - 2(1)/3
c = 7 - 2/3
c = 19/3 = 6 1/3
So the equation is :-
y = 2x/3 + 19/3
Or 3y = 2x + 19

Answer 3

To find the equation of a line, you need to know the gradient and a point.

Gradient of ab = (7-1)/(1-5) = 6/(-4) = -3/2
Gradient perpendicular to ab = -1/(-3/2) = 2/3

Using the point (1,7), substitute the perpendicular gradient into y=mx+c.
7 = (2/3)(1) + c
7 = (2/3) + c
7 - 2/3 = c
c = 6 1/3
c = 19/3

y = (2/3)x + 19/3
3y = 2x + 19

Answer 4

1) work out gradient of line ab:
gradient = (7-1)/(1-5) = -6/4

2) find the negative reciprocal of the gradient:
[-(-6/4)]^(-1) ---> means minus negative six over 4 all to the power of negative one.
this will give you the gradient of the perpendicular line to be 4/6, simplified to be 2/3

3) use point 'a' to find equation of perpendicular line:
7=1(2/3) + c
c= 7-(2/3)
c= 19/3

4) write out the whole equation as your answer:
in the form of y=mx+c will be:
y=(2/3)x + (19/3)

to make things look neater, multiply both sides by 3 (as it is the common denominator on both sides) and rearrange to give the answer in ax+by+c=0:
2x-3y+19=0

Answer 5

first you find the gradient of line ab
gradient of line ab=(1-7)/(5-1) = -3/2

gradient of line perpendicular to ab is therefore 2/3 (since the product of the gradients of two lines perpendicular to each other should be -1)

since you have a point and a gradient of the line passing through that point, you can calculate the eqn of that line

from y=mx+c, where m is the gradient and c the y-intercept,

7=(2/3)(1)+c
c=19/3

eqn is
y=(2/3)x+(19/3)

or 3y=2x+19

Answer 6

gradient of line ab is given by m1,
where m1 =(y2 - y1)/(x2 -x1)
m1 = (1 - 7)/(5 -1) = -6/4 = -3/2
Gradient of perpendicular through a is m2, say.
For perpendicular lines m1 x m2 = -1
Thus -3/2 x m2 = -1
m2 = -1 / (-3/2) = 1 x 2/3 = 2/3
Equation of perpendicular thro`(1,7) is:-
y - 7 = 2/3 (x - 1)
y = (2/3)x -2/3 + 7
y = (2/3)x -2/3 + 21/3
y = (2/3)x + 19/3

Answer 7

m = Slope of a line through (x1, y1) and (x2, y2)
= (y2 - y1) / (x2 - x1)

Slope of ab = m = (1 - 7) / (5 - 1) = -6/4 = - 3/2

Slope of a line perpendicular to ab = -1/m = 2/3

Equation of a line passing through (1, 7): >>{ y - y1 = m(x - x1}

y - 7 = 2/3(x - 1)
Transposing and simplifying
2x - 3y + 19 = 0

Answer 8

y=mx+c.
7 = (2/3)(1) + c
7 = (2/3) + c
7 - 2/3 = c
c = 6 1/3
c = 19/3