Thread: Equation of straight line!

Originally Posted by LoveDeathCab

Hey, i need some help with this question!

9(i) Find the gradient of the line l1 which has the equation 4x-3y+5=0.

I think its 4 by doing this:
4x-3y+5=0
3y=4x+5

y=mx+c where m is the gradient? Is that right?

yes, m is the gradient in y = mx + c ....the equation of a line. So let's find that form. you started but you didn't finish.

4x - 3y + 5 = 0
=> 3y = 4x + 5
=> y = (4/3)x + 5/3 ........divided both sides by 3

Now we can clearly see that the gradient is 4/3

(ii) Find an equation of the line l2 which passes through the point (1,2) and which is perpendicular to the line l1 giving your answer in the form ax+by+c=0

now remember:
(1) two straight lines are parallel if they have the same gradient
(2) two straight lines are perpendicular if their gradients are negative inverses of each other (that means if you take one gradient, turn it upside down by putting 1 over it, and put a minus sign in front, you get the other gradient)

now we want a line perpendicular to l1. the gradient of l1 is 4/3, so the gradient of the line we want is -3/4

using m = -3/4, (x,y) = (1,2), we have:
y = mx + c
=> 2 = (-3/4)(1) + c
=> c = 2 + 3/4 = 11/4

so our line is y = (-3/4)x + 11/4
or (3/4)x + y - 11/4 = 0


Just in case you are familiar with the point-slope form, we could find the equation of the line directly.

using m = -3/4 and (x1,y1) = (1,2), by the point slope form we have:

y - y1 = m(x - x1)
=> y - 2 = (-3/4)(x - 1)
=> y - 2 = (-3/4)x + 3/4
=> (3/4)x + y - 2 - 3/4 = 0
=> (3/4)x + y - 11/4 = 0