Letting L1=ax+by+c=0

Question

and L2=px+qy+r=0.

The equation of anyline passing through the intersection of L1 and L2 is

L1+kL2=0 where k is a variable depending on just how the actual line through the point is.

Can anyone prove this because all i get is 0=0.

Answer 1

Find the point of intersection of the lines in terms of a,b,c,p,q and r.
If I did my math right, the coordinates are
x' = (rb-cq)/(aq-bp)
and
y' = (cp-ar)/(aq-bp)

Any line passing though the intersection can be written
y-y' = m (x - x')
where m is an arbitrary constant.

L1 + kL2 = 0 because of the way L1 and L2 are defined. It is also a general equation for a line
(a + kp)x + (b + kq)y +c + kr = 0

I think the statement proves itself, just by substituting for L1 and L2. By making both substitutions, you require that the equations of both lines are satisfied, so that the equation that results passes through the common point.

Answer 2

but how does

(a + kp)x + (b + kq)y +c + kr = 0

relate to

y-y' = m (x - x')?

Answer 3

It should be possible to rewrite
(a + kp)x + (b + kq)y +c + kr =0 in the other form, with the values of x' and y' that I derived. m will be a function of k and other constants.

y = [(a+kp)/(b+kq)]x + (-c-kr)/(b + kq)

In my first version, the y intercept would be at
y = y' + mx'= (cp-ar)/(aq-bp) + m (rb-cq)/(aq-bp)

I agree they do not look consistent. I may have made an error in the derivation.