# Thread: Geometry - equation of a line

1. ## Geometry - equation of a line

The line L passes through the points A(1,3) and B(-19,-19). Find an equation of L in the form ax+by+c=0.

So the answer in the back of the book is: 11x-10y+19=0

But when I do it:

= -19-3/-19-1
= -16/-18
= 16/18
= 8/9
C =
y=8/9x+c
3=8/9x+c
c=3-8/9
c=27/9-8/9
c=19/9

y=8/9x+19/9

so 8/9x-y+19/9=0
so 8x-9y+19=0

So where did I go wrong?

Im guessing either a stupid mistake in there or the answer in the back of the book is wrong.

2. Originally Posted by Ifailatmaths
The line L passes through the points A(1,3) and B(-19,-19). Find an equation of L in the form ax+by+c=0.

So the answer in the back of the book is: 11x-10y+19=0

But when I do it:

= -19-3/-19-1
= -16/-18
= 16/18
= 8/9

So where did I go wrong?

Im guessing either a stupid mistake in there or the answer in the back of the book is wrong.
You miscalculated the slope in the beginning.

3. Originally Posted by masters
You miscalculated the slope in the beginning.
Cheers new it was a stupid mistake.

= -19-3/-19-1
=-22/-20
=22/20
=11/10

4. You don't really need to calculate the slope. If you given the form ax+ by+ c= 0 and the points (1, 3) and (-19,-19), substitute those x, y values into the equations:
a(1)+ b(3)+ c= a+ 3b+ c= 0 and
a(-19)+ b(-19)+ c= -19a- 19b+ c= 0 and solve for a, b, c.

Yes, that is two equations for three "unknowns" but since you can get the an equation for the same line by mutliplying the entire equation by any number, you can take any one of them, say c, to be any convenient (non-zero) number.

Taking c= 1, a+ 3b= -1 and -19a- 19b= -1. Multiply the first equation by 19 and add equations to eliminate a:
(57- 19)b= 38b= -20 so b= -20/38= -10/19. Then a+ 3(-10/19)= -1, a= 30/19- 1= (30-19)/19= 11/19.

(11/19)x- (10/19)y+ 1= 0 is a perfectly good answer. Of course, you can get rid of the fractions by multiplying the entire equation by 19 to get 11x- 10y+ 19= 0, the answer in your text.