Simple gradient/graph question.

[DeanBH]

1. The points A and b have Coords (6,-1) and (2,5) respectively

a.Show gradient is - 3/2

-1-5 = -6
6-2 = 4

= -3/2

Hence find the equation of the line AB with answer in form Ax + By = C

A B C are integers.

- now this isn't a homework question I am just doing some extra work in my own time. I have never been taught this, but when looking through past papers i have seen that it has actualy come up, can someone explain to me how exactly you do these. I have attempted it, and know the answer because i have the mark scheme, but i don't 100% understand it.


i know you use y-5 and y+1 somewhere, which is the inverse of the Y coordinates in the 2 points I am given. and i know that you somehow times the X by the gradient with those coordinates inverted but I am not sure how it works.

answer = 3x+2y = 16 can anyone explain this?

 

[HallsofIvy]

You got the gradient by (1) subtracting the two y values, (2) subtracting the two x values, (3) dividing the first by the second.

And, of course, you could do that for any two points on the line and get the same thing. Suppose you pick an arbitrary point on the line and call it (x, y). Now use (x,y) and (2, 5) to calculate the gradient. You would have (y- 5)/(x- 2) and that must give you the same thing as (-1-5)/(6-2)= -3/2. That is, you must have (y- 5)/(x-2)= -3/2. "Cross multiplying", that is, multiplying both sides by 2 and (x-2), we get 2(y- 5)= -3(x- 2). Multiply those out, move the "x" term to the left, and the constants to the right and see what you get.

 

[DeanBH]

2y-10=-3x+6
2y+3x=16

hurraynext question is to find equation of perpendicular line that crosses through point b.

is that (y-5)/(X-2) = 2/3

=3y-15 = 2x-4
=3y-2x=11?
or is something wrong